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Commit 3c7c449

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‎.idea/modules.xml

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+15Lines changed: 15 additions & 0 deletions
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‎.idea/vcs.xml

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‎【009】【PalindromeNumber】/src/Solution.java

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/**
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* Author: 王俊超
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* Date: 2015-08-21
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* Time: 16:52
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* Declaration: All Rights Reserved !!!
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*/
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public class Solution {
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/**
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* <pre>
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* 原题
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* Determine whether an integer is a palindrome. Do this without extra space.
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*
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* 题目大意
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* 判断一个数字是否是回访字数，不要使用额外的空间。
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*
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* 解题思路
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* 为了不使用额外的空间，参考了其它的解决，那些解法看起来在isPalindrome方法中没有使用额外参数，
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* 但是却使用了方法调用，这个比一个整数消耗的空间更多 ，并没有达到题目的要求，是假的实现，
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* 所以本题依然采用一个额外的空间进行实现。
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* 首先，负数不是回文数字，其次对数字进行逆转，123变成321这样，如果变换后的数字相等说明是回文数字。
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* </pre>
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*
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* @param x
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* @return
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*/
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public boolean isPalindrome(int x) {
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// 负数不是回访数字
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if (x < 0) {
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return false;
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}
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// 数字逆转后的值，为了不使用溢出采用long
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long reverse = 0;
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int tmp = x;
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// 求逆转后的值
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while (tmp != 0) {
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reverse = reverse * 10 + tmp % 10;
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tmp /= 10;
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}
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// 判断是否是回文数字
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return x == reverse;
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}
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}

‎【009】【PalindromeNumber】/【009】【PalindromeNumber】.iml

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<?xml version="1.0" encoding="UTF-8"?>
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<module type="JAVA_MODULE" version="4">
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<component name="NewModuleRootManager" inherit-compiler-output="true">
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<exclude-output />
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<content url="file://$MODULE_DIR$">
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<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
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</content>
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<orderEntry type="inheritedJdk" />
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<orderEntry type="sourceFolder" forTests="false" />
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</component>
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</module>

‎【014】【LongestCommonPrefix】/src/Solution.java

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/**
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* <pre>
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* 原题
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* 　　Write a function to find the longest common prefix string amongst an array of strings.
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* Write a function to find the longest common prefix string amongst an array of strings.
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*
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* 题目大意
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* 　　写一个函数找出一个字串所数组中的最长的公共前缀。
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* 写一个函数找出一个字串所数组中的最长的公共前缀。
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*
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* 解题思路
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* 　　第一步先找出长度最小的字符串，然后将这个字符串与其它的字符串相比找出最短的最公共前缀。
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* 第一步先找出长度最小的字符串，然后将这个字符串与其它的字符串相比找出最短的最公共前缀。
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* </pre>
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*
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* @param strs

‎【017】【LetterCombinationsOfAPhonNumber 】/src/Solution.java

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import java.util.LinkedList;
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import java.util.List;
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/**
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* Author: 王俊超
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* Date: 2015-08-21
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* Time: 16:23
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* Declaration: All Rights Reserved !!!
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*/
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public class Solution {
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private String[] map = {
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"abc",
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"def",
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"ghi",
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"jkl",
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"mno",
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"pqrs",
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"tuv",
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"wxyz",
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};
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private List<String> result; // 存储最终结果
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private char[] chars; // 保存去掉0，1字符的结果
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private char[] curResult; // 存储中间结果
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private int end = 0; // 字符数组中的第一个未使用的位置
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private int handle = 0; // 当前处理的是第几个字符数字
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/**
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* <pre>
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* 原题
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* Given a digit string, return all possible letter combinations that the number could represent.
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* A mapping of digit to letters (just like on the telephone buttons) is given below.
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*
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* Input:Digit string "23"
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* Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
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*
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* Note: Although the above answer is in lexicographical order, your answer
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* could be in any order you want.
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*
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* 题目大意
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* 给定一个数字串，返回数字上所有字符的所有组合，数字到字符的映射如上图所示。
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* 注意： 尽管上面的结果以字符顺序排列的，你可以以任何顺序返回结果。
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*
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* 解题思路
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* 用一个数组保存数字和字的映射关系，根据数字串的输入，找到对应的字符，组合结果。
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* </pre>
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*
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* @param digits
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* @return
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*/
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public List<String> letterCombinations(String digits) {
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result = new LinkedList<>();
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if (digits != null && digits.length() > 0) {
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chars = digits.toCharArray();
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// 对字符串进行处理，去掉0和1
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// 找第一个0或者1的位置
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while (end < digits.length() && chars[end] != '0' && chars[end] != '1') {
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end++;
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}
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handle = end + 1;
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while (handle < chars.length) {
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if (chars[handle] != '0' && chars[handle] != '1') {
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chars[end] = chars[handle];
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end++;
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}
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handle++;
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}
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curResult = new char[end];
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// while结束后，end为有效字符的长度
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handle = 0; // 指向第一个有效字符的位置
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letterCombinations();
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}
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return result;
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}
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private void letterCombinations() {
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if (handle >= end) {
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result.add(new String(curResult));
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} else {
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int num = chars[handle] - '2';
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for (int i = 0; i < map[num].length(); i++) {
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curResult[handle] = map[num].charAt(i);
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handle++;
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letterCombinations();
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handle--;
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}
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}
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}
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}

‎【017】【LetterCombinationsOfAPhonNumber 】/src/phone-number.png

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‎【017】【LetterCombinationsOfAPhonNumber 】/【017】【LetterCombinationsOfAPhonNumber 】.iml

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<?xml version="1.0" encoding="UTF-8"?>
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<module type="JAVA_MODULE" version="4">
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<component name="NewModuleRootManager" inherit-compiler-output="true">
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<exclude-output />
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<content url="file://$MODULE_DIR$">
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<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
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</content>
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<orderEntry type="inheritedJdk" />
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<orderEntry type="sourceFolder" forTests="false" />
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</component>
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</module>

‎【018】【4Sum】/src/Solution.java

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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.LinkedList;
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import java.util.List;
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/**
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* Author: 王俊超
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* Date: 2015-08-21
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* Time: 16:28
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* Declaration: All Rights Reserved !!!
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*/
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public class Solution {
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/**
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* <pre>
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* 原题
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* Given an array S of n integers, are there elements a, b, c, and d in S
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* such that a + b + c + d = target? Find all unique quadruplets in the array
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* which gives the sum of target.
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* Note:
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* Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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* The solution set must not contain duplicate quadruplets.
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*
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* For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
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*
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* A solution set is:
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* (-1, 0, 0, 1)
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* (-2, -1, 1, 2)
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* (-2, 0, 0, 2)
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*
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* 题目大意
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* 给定一个整数数组，找出a + b + c + d = target的唯一解。
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*
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* 解题思路
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* 先确定a和d的两个数，对于a和d两个数，不能同时重复使用。然后再确定b和c，同样这两个数也不能
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* 同时重复使用。找出所有满足条件的解，同时可以保证解不重复。
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* </pre>
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*
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* @param num
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* @param target
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* @return
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*/
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public List<List<Integer>> fourSum(int[] num, int target) {
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List<List<Integer>> result = new LinkedList<>();
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if (num == null || num.length < 4) {
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return result;
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}
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Arrays.sort(num); // 对数组进行排序
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for (int i = 0; i < num.length - 3; i++) { // 第一个加数
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if (i > 0 && num[i] == num[i - 1]) { // 第一个加数使用不重复
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continue;
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}
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for (int j = num.length - 1; j > i + 2; j--) { // 第四个加数
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if (j < num.length - 1 && num[j] == num[j + 1]) { // 第四个加数使用不重复
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continue;
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}
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int start = i + 1; // 第二个加数
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int end = j - 1; // 第三个加数
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int n = target - num[i] - num[j];
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while (start < end) {
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if (num[start] + num[end] == n) {
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List<Integer> four = new ArrayList<>(4);
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four.add(num[i]);
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four.add(num[start]);
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four.add(num[end]);
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four.add(num[j]);
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result.add(four);
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do {
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start++;
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} while (start < end && num[start] == num[start - 1]); // 保证再次使用第二个数不重复
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do {
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end--;
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} while (start < end && num[end] == num[end + 1]); // 保证再次使用第三个数不重复
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} else if (num[start] + num[end] < n) {
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do {
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start++;
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} while (start < end && num[start] == num[start - 1]); // 保证再次使用第二个数不重复
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} else {
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do {
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end--;
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} while (start < end && num[end] == num[end + 1]); // 保证再次使用第三个数不重复
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}
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}
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}
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}
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return result;
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}
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}

‎【018】【4Sum】/【018】【4Sum】.iml

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<?xml version="1.0" encoding="UTF-8"?>
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<module type="JAVA_MODULE" version="4">
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<component name="NewModuleRootManager" inherit-compiler-output="true">
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<exclude-output />
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<content url="file://$MODULE_DIR$">
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<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
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</content>
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<orderEntry type="inheritedJdk" />
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<orderEntry type="sourceFolder" forTests="false" />
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</component>
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</module>

‎【019】【RemoveNthNodeFromEndOfList】/src/ListNode.java

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/**
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* Author: Íõ¿¡³¬
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* Date: 2015-08-21
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* Time: 16:30
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* Declaration: All Rights Reserved !!!
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*/
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public class ListNode {
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int val;
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ListNode next;
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ListNode(int x) {
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val = x;
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next = null;
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}
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}

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