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FlipInteger.java
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86 lines (74 loc) · 2.37 KB
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package LeetCodeOJ;
import offer.constructArray52;
/**
* leetcode 476 Number Complement Given a positive integer, output its
* complement number. The complement strategy is to flip the bits of its binary
* representation. Example 1: Input: 5 Output: 2 Explanation: The binary
* representation of 5 is 101 (no leading zero bits), and its complement is 010.
* So you need to output 2. Example 2: Input: 1 Output: 0 Explanation: The
* binary representation of 1 is 1 (no leading zero bits), and its complement is
* 0. So you need to output 0.
*
* @author fqx
*
*/
public class FlipInteger {
public static int findComplement(int n) {
String a = Integer.toBinaryString(n);
int i = 0;
for (i = 0; i < a.length() - 1; i++) {
if (a.charAt(i) == '0') {
continue;
} else {
break;
}
}
char[] ch = a.substring(i).toCharArray();
for (int j = 0; j < ch.length; j++) {
if (ch[j] == '0') {
ch[j] = '1';
} else {
ch[j] = '0';
}
}
int result = Integer.parseInt(new String(ch), 2);
return result;
}
/*
* 别人的算法
*/
public int findComplement_1(int num) {
if (num == 0) {
return 0;
}
//正数左移 直到 等于0 说明，非零部分全部移出。
int count = 0;
while (num > 0) {
num = num << 1;
count++;
}
//只截取正数中从高位开始碰到第一个1的后面的所有二进制
num = num >> count;
//负数 直接取反，因为负数存储的是补码 ，高位全部shi
num = ~num;
return num;
}
/*
* 计算机天才们的想法
*/
public int findComplete_2(int num){
int leftmost = Integer.highestOneBit(num);
int mask = ( leftmost << 1) - 1;
num = ~num;
return num & mask;
/* int mask = 1;
while (mask < num) mask = (mask << 1) | 1;
return ~num & mask;*/
}
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(findComplement(-5));
System.out.println(new FlipInteger().findComplement_1(-5));
System.out.println(new FlipInteger().findComplete_2(-5));
}
}