|
| 1 | +/* |
| 2 | + Author: King, wangjingui@outlook.com |
| 3 | + Date: Dec 17, 2014 |
| 4 | + Problem: Implement strStr() |
| 5 | + Difficulty: Easy |
| 6 | + Source: https://oj.leetcode.com/problems/implement-strstr/ |
| 7 | + Notes: |
| 8 | + Implement strStr(). |
| 9 | + Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. |
| 10 | +
|
| 11 | + Solution: 1. Check in the haystack one by one. If not equal to needle, reset the pointer.(TLE) |
| 12 | + 2. Classice KMP solution. |
| 13 | + 3. Simplified RK Soluiton. Thanks for [wenyuanhust, wenyuanhust@gmail.com] |
| 14 | + */ |
| 15 | + |
| 16 | +public class Solution { |
| 17 | + public int strStr_1(String haystack, String needle) { |
| 18 | + int sLen = haystack.length(), tLen = needle.length(); |
| 19 | + if(tLen == 0) return 0; |
| 20 | + if (haystack==null || needle==null || sLen==0) return -1; |
| 21 | + int i = 0, j = 0; |
| 22 | + while (i < sLen) { |
| 23 | + for (j = 0; j < tLen && (i+j) < sLen && haystack.charAt(i+j) == needle.charAt(j); ++j); |
| 24 | + if (j == tLen) return i; |
| 25 | + ++i; |
| 26 | + } |
| 27 | + return (int)-1; |
| 28 | + } |
| 29 | + void getNext(String T, int[] next){ |
| 30 | + int i=0, j=-1; |
| 31 | + next[0]=-1; |
| 32 | + int n = next.length; |
| 33 | + while(i < n - 1){ |
| 34 | + while(j>-1&&T.charAt(j)!=T.charAt(i)) j = next[j]; |
| 35 | + ++i; ++j; |
| 36 | + if(i < n - 1 && j < n - 1 && T.charAt(j)==T.charAt(i)) next[i]=next[j]; |
| 37 | + else next[i]=j; |
| 38 | + } |
| 39 | + } |
| 40 | + public int strStr_2(String haystack, String needle) { |
| 41 | + int sLen = haystack.length(), tLen = needle.length(); |
| 42 | + if(tLen == 0) return 0; |
| 43 | + if (haystack==null || needle==null || sLen==0) return -1; |
| 44 | + int[] next = new int[tLen+1]; |
| 45 | + getNext(needle, next); |
| 46 | + int i = 0, j = 0; |
| 47 | + while (i < sLen) { |
| 48 | + while (j > -1 && needle.charAt(j) != haystack.charAt(i)) j = next[j]; |
| 49 | + ++i; ++j; |
| 50 | + if (j == tLen) return i - j; |
| 51 | + } |
| 52 | + return -1; |
| 53 | + } |
| 54 | + public int strStr_3(String haystack, String needle) { |
| 55 | + int sLen = haystack.length(), tLen = needle.length(); |
| 56 | + if (tLen == 0) return 0; |
| 57 | + if (haystack==null || needle==null || sLen==0 || sLen < tLen) return -1; |
| 58 | + long fh = 0, fn = 0; |
| 59 | + int head = 0, tail = tLen - 1; |
| 60 | + for (int i = 0; i < tLen; ++i) { |
| 61 | + fh += haystack.charAt(i); |
| 62 | + fn += needle.charAt(i); |
| 63 | + } |
| 64 | + while (tail < sLen) { |
| 65 | + if (fn == fh) { |
| 66 | + int i = 0; |
| 67 | + while (i < tLen && needle.charAt(i) == haystack.charAt(i + head)){ |
| 68 | + ++i; |
| 69 | + } |
| 70 | + if (i == tLen) return head; |
| 71 | + } |
| 72 | + if (tail == sLen - 1) return -1; |
| 73 | + fh -= haystack.charAt(head++); |
| 74 | + fh += haystack.charAt(++tail); |
| 75 | + } |
| 76 | + return -1; |
| 77 | + } |
| 78 | +} |
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