LeetCode OJ 476 Number Complement [Easy]

题目描述：

Given a positiveinteger, output its complement number. The complement strategy is to flip thebits of its binary representation.

Note:

The given integer isguaranteed to fit within the range of a 32-bit signed integer.

You could assume noleading zero bit in the integer’s binary representation.

Example 1:

Input: 5

Output: 2

Explanation: Thebinary representation of 5 is 101 (no leading zero bits), and its complement is010. So you need to output 2.

Example 2:

Input: 1

Output: 0

Explanation: Thebinary representation of 1 is 1 (no leading zero bits), and its complement is0. So you need to output 0.

题目理解：

给定一个正整数，输出它的补码。补码是将该正整数的二进制全部翻转后得到的整数

我的分析：

1.  依次，用2除，求余数，求商；

2.  由此得到二进制的反序列；

3.  将序列放入整型数组，再依次取出翻转求出补码。

我的解答：

public static int findComplement(int num){        int result = 0;        int [] bitnum = new int[32];        int range = 0;        for(int i = 0; num != 0; i++){            bitnum[i] = num % 2;            num /= 2;            range = i;        }        for(int i = 0; i <= range; i++){            if(bitnum[i] == 0) result += (bitnum[i] + 1) * Math.pow(2, i);            else result += (bitnum[i] - 1) * Math.pow(2, i);        }        return result;    }

其他的解答1：

//100110, its complement is 011001, the sum is 111111. So we only need get the min number large or equal to num, then do substractionpublic int findComplement(int num) {    int i = 0;    int j = 0;    while (i < num) {        i += Math.pow(2, j);        j++;    }    return i - num;}

对解答1的改动：

一个数乘2，相当于左移一位

//Same idea, but using bit manipulation instead of Math.pow().public class Solution {    public int findComplement(int num) {        int n = 0;        while (n < num) {            n = (n << 1) | 1;        }        return n - num;    }}

更快的解答1：

import java.lang.*;public class Solution {    public int findComplement(int num) {        int counter = (int) (Math.floor((Math.log(num)/Math.log(2))) + 1);        return (((~num) << (32-counter)) >>> (32-counter));    }}

更快的解答2：

public class Solution {    public int findComplement(int num) {        return num ^ ((Integer.highestOneBit(num) << 1) - 1);    }}