[LeetCode] 476. Number Complement

[grandyang]

 

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

Note:

1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
2. You could assume no leading zero bit in the integer’s 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.

 

这道题给了我们一个数，让我们求补数。通过分析题目汇总的例子，我们知道需要做的就是每个位翻转一下就行了，但是翻转的起始位置上从最高位的1开始的，前面的0是不能被翻转的，所以我们从高往低遍历，如果遇到第一个1了后，我们的flag就赋值为true，然后就可以进行翻转了，翻转的方法就是对应位异或一个1即可，参见代码如下：

 

解法一：

class Solution {
public:
    int findComplement(int num) {
        bool start = false;
        for (int i = 31; i >= 0; --i) {
            if (num & (1 << i)) start = true;
            if (start) num ^= (1 << i);
        }
        return num;
    }
};

 

由于位操作里面的取反符号～本身就可以翻转位，但是如果直接对num取反的话就是每一位都翻转了，而最高位1之前的0是不能翻转的，所以我们只要用一个mask来标记最高位1前面的所有0的位置，然后对mask取反后，与上对num取反的结果即可，参见代码如下：

 

解法二：

class Solution {
public:
    int findComplement(int num) {
        int mask = INT_MAX;
        while (mask & num) mask <<= 1;
        return ~mask & ~num;
    }
};

 

再来看一种迭代的写法，一行搞定碉堡了，思路就是每次都右移一位，并根据最低位的值先进行翻转，如果当前值小于等于1了，就不用再调用递归函数了，参见代码如下：

 

解法三：

class Solution {
public:
    int findComplement(int num) {
        return (1 - num % 2) + 2 * (num <= 1 ? 0 : findComplement(num / 2));
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/74627/3-line-c

https://discuss.leetcode.com/topic/74968/simple-java-one-line-solution

https://discuss.leetcode.com/topic/74642/java-1-line-bit-manipulation-solution

 

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