1. 题目

给定一个32位整数 num，你可以将一个数位从0变为1。请编写一个程序，找出你能够获得的最长的一串1的长度。

示例 1：
输入: num = 1775(11011101111)
输出: 8示例 2：
输入: num = 7(0111)
输出: 4

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-bits-lcci
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 记录连续1的端点及长度

class Solution {
public:int reverseBits(int num) {if(num==0)return 1;int prevlen, prevEnd, maxlen = 0, count = 0, start=-1;int curstart, curlen, curEnd;map<int,pair<int,int>> continOne;//连续1，结束pos ： {开始位置，个数}for(int i = 0; i < 32; ++i){if((num>>i)&1) //为1{if(start==-1)start = i;count++;if(i == 31)continOne[i] = make_pair(start,count);}else{if(count)continOne[i-1] = make_pair(start,count);count = 0;start = -1;}}auto it = continOne.begin();prevlen = it->second.second;prevEnd = it->first;maxlen = prevlen+1;it++;for(; it != continOne.end(); ++it){curEnd = it->first;curstart = it->second.first;curlen = it->second.second;if(curstart - prevEnd == 2)maxlen = max(maxlen, curlen+prevlen+1);elsemaxlen = max(maxlen, curlen+1);prevlen = curlen;prevEnd = curEnd;}return maxlen;}
};

• 简洁版

class Solution {
public:int reverseBits(int num) {int prevlen = 0, curlen = 0, maxlen = 0;for(int i = 0; i < 32; ++i){if((num>>i)&1) //为1curlen++;else{maxlen = max(maxlen, curlen+prevlen+1);prevlen = curlen;curlen = 0;}}return maxlen;}
};