LeetCode 868 Binary Gap

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Description

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: 22
Output: 2
Explanation: 
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: 8
Output: 0
Explanation: 
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

• 解题思路

   根据题目，是要求二进制数相邻两个1的最大间距，如果没有两个相邻的1则返回0。由1 <= N <= 10^9 可知N是一个32位的整数，使用移位，从后至前查找最大间距。

• AC代码

class Solution {
public:
    int binaryGap(int N) {
        int dist = 0, flag = -1;
        for (int i = 0; i < 32; i++) {
            if ((N & (1 << i)) != 0) {
                if (flag != -1) dist = max(dist, i - flag);
                flag = i;
            }
        }
        return dist;
    }
};