2022年01月18日力扣每日一题

题目

给定一个 24 小时制（小时:分钟 "HH:MM"）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：

输入：timePoints = ["23:59","00:00"]
输出：1

示例 2：

输入：timePoints = ["00:00","23:59","00:00"]
输出：0

提示：

2 <= timePoints.length <= 2 * 10⁴
timePoints[i] 格式为 "HH:MM"

个人解法

import java.util.List;

class Solution {
    public int findMinDifference(List<String> timePoints) {
        int[] times = new int[2880];
        for (String timePoint : timePoints) {
            String[] strs = timePoint.split(":");
            int time = Integer.parseInt(strs[0]) * 60 + Integer.parseInt(strs[1]);
            if (times[time] == 1) {
                return 0;
            }
            times[time] = 1;
            times[time + 1440] = 1;
        }
        if (times[0] == 1 && times[1439] == 1) {
            return 1;
        }
        int min = 1440;
        int bef = 0;
        for (int i = 1; i < 2880; i++) {
            if (times[i] == 1) {
                if (bef > 0) {
                    min = Math.min(min, i - bef);
                }
                if (i > 1439) {
                    break;
                }
                bef = i;
            }
        }
        return min;
    }
}

from typing import List


class Solution:
    def findMinDifference(self, timePoints: List[str]) -> int:
        times = [0] * 2880
        for timePoint in timePoints:
            time = int(timePoint[:2]) * 60 + int(timePoint[-2:])
            if times[time] == 1:
                return 0
            times[time] = 1
            times[time + 1440] = 1
        result = 1440
        bef = 0
        for i in range(2880):
            if times[i] == 1:
                if bef > 0:
                    result = min(result, i - bef)
                if i > 1439:
                    break
                bef = i
        return result