476. Number Complement

题目：

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.

链接：https://leetcode.com/problems/number-complement/#/description

3/28/2017

做完一遍要立即差错。

public class Solution {
    public int findComplement(int num) {
        int tmp = num ^ 0xffffffff;
        int reset = 0x80000000;
        for (int i = 0; i < 32; i++) {
            if ((tmp & reset) != 0) {
                tmp &= (reset ^ 0xffffffff);
                reset >>= 1;
            }
            else break;
        }
        return tmp;
    }
}

别人的好算法，右半部分是mask，把最右边的位置1。看来要系统研究一下Java本身的一些方法，比如Integer的很多跟bit有关的方法

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

C++，mask右边补0，最后取反也跟前面的mask一样了

class Solution {
public:
    int findComplement(int num) {
        unsigned mask = ~0;
        while (num & mask) mask <<= 1;
        return ~mask & ~num;
    }
};

这个方法也很有意思，是将从最高的为1的位开始，所有右边的位数置为1，每次重复都是2倍的长度。原因？最左边位是1，右移之后原来此左边位也变成了1，第二次时候把最高的2位都用来置右边2位。直到最后一步为总位数的一半

int findComplement(int num) {
    int mask = num;
    mask |= mask >> 1;
    mask |= mask >> 2;
    mask |= mask >> 4;
    mask |= mask >> 8;
    mask |= mask >> 16;
    return num ^ mask;
}

更多讨论：https://discuss.leetcode.com/category/608/number-complement