[LeetCode] 868. Binary Gap

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原题链接： https://leetcode.com/problems/binary-gap/

1. 题目介绍

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.
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 。

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

2. 解题思路

注：本题思路参考了 https://www.cnblogs.com/king-3/p/9332944.html 的内容，感兴趣的读者可以去看原文。
首先寻找第一个1，记录第一个1的位置。
然后寻找其他的1，在寻找的过程中逐步更新最长的距离。

实现代码

class Solution {
    public int binaryGap(int N) {

		int Count = 0;//记录位置
		int distance = 0;//需要返回的距离

        //寻找第一个1的位置
		int firstOne = 0;//第一个1的位置
		while(N != 0) {
			Count ++;
			int k = N & 1;
        	if (k == 1) {
        		firstOne = Count;
        		N = N>>1;
        		break;
        	}
        	N = N>>1;
		}
        
        //寻找其他的1，并且更新最长的距离
		while(N != 0){
			Count++;
			int k = N & 1;
        	if (k == 1) {
        		distance = Math.max(distance , Count - firstOne);
        		firstOne = Count;
        	}
        	N = N>>1;
		}
		return distance;
    }
}