Description

Difficulty：Easy

Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.

If there aren't two consecutive 1's, return 0.

Example 1:

Input: 22Output: 2Explanation: 22 in binary is 0b10110.In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.The first consecutive pair of 1's have distance 2.The second consecutive pair of 1's have distance 1.The answer is the largest of these two distances, which is 2.

Example 2:

Input: 5Output: 2Explanation: 5 in binary is 0b101.

Example 3:

Input: 6Output: 1Explanation: 6 in binary is 0b110.

Example 4:

Input: 8Output: 0Explanation: 8 in binary is 0b1000.There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.

Note:

1 <= N <= 10^9

Solution

其实是一道数1的题,用 two pointers 解决

func binaryGap(N int) int {    max,pre,next := 0,0,1    for  ;N & 1 == 0 && N > 0; pre,N = pre+1, N >> 1{    }        for next,N = pre + 1,N>>1;N > 0; N>>=1 {        for N & 1 == 0{            next++            N >>=1        }        if next -pre > max {            max = next - pre        }        pre = next        next += 1    }        return max}

网友有一个更精妙简单的：

public int binaryGap(int N) {        int res = 0;        for (int d = -32; N > 0; N /= 2, d++)            if (N % 2 == 1) {                res = Math.max(res, d);                d = 0;            }        return res;    }

d 是两个1之间距离，碰到1之后d置0.