classSolution {public:intfindMinDifference(vector<string>&timePoints) {intres = INT_MAX, n = timePoints.size(), diff =0; sort(timePoints.begin(), timePoints.end());for(inti =0; i < n; ++i) {stringt1 = timePoints[i], t2 = timePoints[(i +1) %n];inth1 = (t1[0] -'0')...
:rtype: int """defconvert(time):returnint(time[:2]) *60+int(time[3:]) timePoints =map(convert, timePoints) timePoints.sort()returnmin((y - x) % (24*60)forx, yinzip(timePoints, timePoints[1:] + timePoints[:1])) 参考资料:https://leetcode.com/problems/minimum-time-difference...
LeetCode力扣 410. 分割数组的最大值 Split Array Largest Sum 110 -- 10:39 App LeetCode力扣 994. 腐烂的橘子 Rotting Oranges 112 -- 9:07 App LeetCode力扣 509. 斐波那契数 Fibonacci Number 38 -- 7:33 App LeetCode力扣 153. 寻找旋转排序数组中的最小值 Find Minimum in Rotated Sorted 130...
classSolution { public:intfindMinDifference(vector<string>&timePoints) {intres = INT_MAX, pre =0, first = INT_MAX, last =INT_MIN; vector<int> mask(1440,0);for(stringstr : timePoints) {inth = stoi(str.substr(0,2)), m = stoi(str.substr(3));if(mask[h *60+ m] ==1)return0...
My second solution(44ms) 代码语言:javascript 代码运行次数:0 运行 AI代码解释 publicintfindMinDifference(List<String>timePoints){int[]times=newint[timePoints.size()];int index=0;for(String num:timePoints){String[]time=num.split(":");int minutes=Integer.parseInt(time[0])*60+Integer.parseInt...
class Solution { public: int findMinDifference(vector<string>& timePoints) { int res=INT_MAX; int n=timePoints.size(); sort(timePoints.begin(),timePoints.end()); for(int i=0;i<n;i++){ string t1=timePoints[i]; string t2=timePoints[(i+1)%n]; ...
Time Complexity: O(timePoints.size() + 24*60). Space: O(24*60). AC Java: AI检测代码解析 1classSolution {2publicintfindMinDifference(List<String>timePoints) {3boolean[] mark =newboolean[24*60];4for(String s : timePoints){5String [] parts = s.split(":");6inttime = Integer.valu...
539. Minimum Time Difference JHY还能卷 我偏要勉强。 来自专栏 · LeetCode刷题记录 539. Minimum Time Difference 难度:medium Given a list of 24-hour clock time points in "Hour:Minutes" format, find the minimumminutesdifference between any two time points in the list. 这道题总体比较简单,只需...
[LeetCode] Find Minimum in Rotated Sorted Array II 简介:This problem is more or less the same as Find Minimum in Rotated Sorted Array. And one key difference is as stated in the solution tag. This problem is more or less the same asFind Minimum in Rotated Sorted Array. And one key...
More practice:If you have figured out theO(n) solution, try coding another solution of which the time complexity isO(nlogn). Approach#1 Brute Force Brute force方法就是两个循环,start和end指针之间的数加起来。复杂度O(n2),一次AC。没毛病,老铁。