Force deduction solution summary 868- binary spacing

Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

Given a positive integer n, Find and return to n In the binary representation of adjacent 1 Between Longest distance . If there are no two adjacent 1, return 0 .

If only 0 Put two 1 Separate （ May not exist 0 ）, We think these two 1 each other adjacent . Two 1 The distance between them is the absolute difference of their positions in the binary representation . for example ,"1001" Two of them 1 The distance to 3 .

Example 1：

Input ：n = 22
Output ：2
explain ：22 The binary of is "10110" .
stay 22 In the binary representation of , There are three 1, Make up two adjacent pairs of 1 .
The first pair of adjacent 1 in , Two 1 The distance between is 2 .
The second pair is adjacent 1 in , Two 1 The distance between is 1 .
The answer is the largest of the two distances , That is to say 2 .
Example 2：

Input ：n = 8
Output ：0
explain ：8 The binary of is "1000" .
stay 8 There are no adjacent two in the binary representation of 1, So back 0 .
Example 3：

Input ：n = 5
Output ：2
explain ：5 The binary of is "101" .
 

Tips ：

1 <= n <= 109

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/binary-gap
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  To string, Then judge the distance 

Code ：

public class Solution868 {

    public int binaryGap(int n) {
        char[] chars = Integer.toString(n, 2).toCharArray();
        int result = 0;
        int currentIndex = 0;
        for (int i = 0; i < chars.length; i++) {
            char aChar = chars[i];
            if (aChar == '0') {
                continue;
            }
            result = Math.max(i - currentIndex, result);
            currentIndex = i;
        }

        return result;
    }
}