leetcode 539. Minimum Time Difference 时间最小差值 + C++中stringstream一个很好示例

末蓝、 2022-06-03 03:39 115阅读 0赞

Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.  
Example 1:  
Input: \[“23:59”,”00:00”\]  
Output: 1  
Note:  
The number of time points in the given list is at least 2 and won’t exceed 20000.  
The input time is legal and ranges from 00:00 to 23:59.

题意很简单，注意这几个事项：  
1）对string直接排序即可，字典排序是可以的  
2）对于最后一个时间节点的计算需要加上24\*60min

同样的本题注意对C++的stringstream的使用

建议和这一道题[leetcode 537. Complex Number Multiplication C++中的stringstream真的很好用][leetcode 537. Complex Number Multiplication C_stringstream] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    class Solution 
    {
    public:
        int findMinDifference(vector<string>& timePoints) 
        {
            int minDif = 24 * 60;
            sort(timePoints.begin(),timePoints.end());
            for (int i = 0; i < timePoints.size(); i++)
            {
                string a = timePoints[i], b = timePoints[(i + 1) % timePoints.size()];
                stringstream sa(a), sb(b);
                int h1, m1, h2, m2;
                char c;
                sa >> h1 >> c >> m1;
                sb >> h2 >> c >> m2;
                int dif = (h2 - h1) * 60 + (m2 - m1);
                if (i == timePoints.size() - 1)
                    dif += 24 * 60;
                minDif = min(minDif, dif);
            }
            return minDif;
        }
    };

[leetcode 537. Complex Number Multiplication C_stringstream]: http://blog.csdn.net/jackzhang_123/article/details/78825050