> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/868.binary_gap.md).

# 868.Binary Gap

**868.Binary Gap**

难度:Easy

> 给定一个正整数 N，找到并返回 N 的二进制表示中两个连续的 1 之间的最长距离。

如果没有两个连续的 1，返回 0 。

```
示例 1：

输入：22
输出：2
解释：
22 的二进制是 0b10110 。
在 22 的二进制表示中，有三个 1，组成两对连续的 1 。
第一对连续的 1 中，两个 1 之间的距离为 2 。
第二对连续的 1 中，两个 1 之间的距离为 1 。
答案取两个距离之中最大的，也就是 2 。
示例 2：

输入：5
输出：2
解释：
5 的二进制是 0b101 。
示例 3：

输入：6
输出：1
解释：
6 的二进制是 0b110 。
示例 4：

输入：8
输出：0
解释：
8 的二进制是 0b1000 。
在 8 的二进制表示中没有连续的 1，所以返回 0 。
 

提示：

1 <= N <= 10^9
```

基本的位运算:

```
class Solution {
public:
    int binaryGap(int N) {
        int k=0;
        int res=0;
        int flag=0;
        while(N)
        {
            if(N&1)
            {
                 if(flag) 
                     res=max(res,k);
                else                 
                    flag=1; 
                  k=1;        
            }
            else if(flag)
                k++;
            N= N>>1;                             
            //cout<<N<<"and"<<k<<endl;
        }
        return res;
    }
};
```

> 执行用时 : 8 ms, 在Binary Gap的C++提交中击败了94.64% 的用户 内存消耗 : 8.2 MB, 在Binary Gap的C++提交中击败了76.85% 的用户


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