[LintCode] Flip Bits

joyvw 发布于2019-08-14 15:21 / 885人阅读

摘要：的二进制补码就是个，因此这道题一定要考虑正负号的问题。然后去检查的二进制包含多少个，方法是对的每一位除以取余。如果为，就说明那一位为，即和在那一位不同，需要进行转换。每次取余之后，减小为二分之一，删除已经检查过的高位。

Problem

Determine the number of bits required to flip if you want to convert integer n to integer m.

Example

Given n = 31 (11111), m = 14 (01110), return 2.

Note

-1的二进制补码就是32个1，因此这道题一定要考虑正负号的问题。
先将a和b异或得到c，若c为负，就说明a和b的符号不同，所以要多一次符号位的转换。
然后去检查c的二进制包含多少个1，方法是对c的每一位除以2取余。如果为1，就说明那一位为1，即a和b在那一位不同，需要进行转换。每次取余之后，c减小为二分之一，删除已经检查过的高位。

Solution

class Solution {
    public static int bitSwapRequired(int a, int b) {
        int c = a ^ b, count = 0;
        if (c < 0) {
            c ^= Integer.MIN_VALUE;
            count++;
        }
        while (c != 0) {
            count += c % 2;
            c /= 2;
        }
        return count;
    }
}

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