LeetCode - 868. Binary Gap

最新推荐文章于 2021-05-13 22:42:15 发布

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最新推荐文章于 2021-05-13 22:42:15 发布

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原文链接：http://www.cnblogs.com/wxisme/p/9561674.html

博客围绕给定正整数，探讨如何找出其二进制表示中两个连续1之间的最长距离，若不存在两个连续1则返回0，还给出了示例，内容与算法相关。

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

Example 3:

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

Example 4:

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

class Solution {
    public int binaryGap(int N) {
        int cnt = -32, ret = 0;
        while (N > 0) {
            if ((N & 1) == 1) {
                ret = Math.max(ret, cnt);
                cnt = 0;
            }
            N = N >> 1;
            cnt++;
        }
        return ret;
    }
}

转载于:https://www.cnblogs.com/wxisme/p/9561674.html