leetcode 681. Next Closest Time Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused. You may assume the given input string is always valid. For example, "01:34", "12:...
return time[:4] + c4 c3 = self.findNext(time[3], nums) if c3 and int(c3 + nums[0]) < 60: return time[:3] + c3 + nums[0] c1 = self.findNext(time[1], nums) if c1 and int(time[0] + c1) <= 24: return time[0] + c1 + ':' + nums[0] + nums[0] ...
Use new time hash value - current time hash value and maintain the minimum diff. If minused result < 0, it is next day, add it with 24 * 60. Time Complexity: O(1). 共有4^4种组合. Space: O(1). size为4的HashSet. AC Java: 1 class Solution { 2 public String nextClosestTime(...
LeetCode 681. Next Closest Time 原题链接在这里:https://leetcode.com/problems/next-closest-time/ 题目: Given a time represented in the format "HH:MM", form the next closest time by reusing the LeetCode String git sed java 转载 mob604756fbd94e 2017-10-22 05:34:00 232阅读 2评论 ...
classSolution{publicStringnextClosestTime(Stringtime){Set<Integer>allowed=newHashSet<>();for(inti=0;i<time.length();++i){if(time.charAt(i)!=':'){// don't forget to exclude ':'allowed.add(time.charAt(i)-'0');}}String[]arr=time.split(":");intsecond=60*Integer.parseInt(arr[0]...
leetcode 1266. Minimum Time Visiting All Points 2019-11-24 21:40 − On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit ... 琴影 0 624 golang time json mongodb 时间处理 2019-12-15 09:...
681. Next Closest Time Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused. You may assume the given input string is always valid. For example, "01:34", "12:09" are...
public class Solution { /** * @param time: the given time * @return: the next closest time */ public String nextClosestTime(String time) { // write your code here char tTime[] = time.toCharArray(); int[] number = new int[]{tTime[0]-'0', tTime[1]-'0', tTime[3]-'0',...
原题链接在这里:https://leetcode.com/problems/next-closest-time/ 题目: Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused. ...
Time Complexity: O(x), x是n的digit 数目. 每个digit不会走超过3遍. Space: O(x). AC Java: AI检测代码解析 1 class Solution { 2 public int nextGreaterElement(int n) { 3 char [] arr = (""+n).toCharArray(); 4 int i = arr.length-2; ...