LeetCode: 868. Binary Gap

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本文解析了LeetCode上编号为868的问题Binary Gap，该问题要求找出给定正整数N的二进制表示中两个连续1之间的最大距离。文章详细解释了解题思路，并给出了C++实现的示例代码。

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LeetCode: 868. Binary Gap

题目描述

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

解题思路

这是签到题，直接计算转换为二进制数后的相邻 1 的距离，记录距离最远的 1 的距离。

AC 代码

class Solution {
public:
    int binaryGap(int N) {
        int curPos = 0;
        int lastOnePos = -1;
        int maxDis = 0;

        while(N)
        {
            if(N%2 == 1)
            {
                if(lastOnePos != -1)
                {
                    maxDis = max(curPos - lastOnePos, maxDis);
                }
                lastOnePos = curPos;
            }

            N /= 2;
            ++curPos;
        }

        return maxDis;
    }
};