LeetCode-Java-868. Binary Gap

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题目

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

代码

此方法其实是每次找到1的时候就记录一次,然后每次看上一次记录的和这次记录的那个差值比较大，留下大的((N >> i) & 1) > 0这段代码就是找1的过程，至于为什么循环次数是32是因为正整数的范围为[0,2^31-1]所以只需要右移32次，此数肯定已经变成了0，i-last中i为当前的位置而last为上一次出现1的位置，想减就可以求得其两个1之间的距离

class Solution {
    public int binaryGap(int N) {
        int last = -1, ans = 0;
        for (int i = 0; i < 32; ++i){
            if (((N >> i) & 1) > 0) {
                if (last >= 0)
                    ans = Math.max(ans, i - last);
                last = i;
            }
        }
        return ans;
    }
}