Leetcode 1629. Slowest Key

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

解析：Version 1，遍历每个字符，计算其时间间隔，如果比之前的大，则更新时间间隔及字符，如果相等，则比较字符大小，判断是否更新字符。

• Version 1

class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        n = len(keysPressed)
        key = keysPressed[0]
        time = releaseTimes[0]
        for i in range(1, n):
            temp = releaseTimes[i] - releaseTimes[i-1]
            if temp > time:
                key = keysPressed[i]
                time = temp
            elif temp == time and keysPressed[i] > key:
                key = keysPressed[i]
        return key

Reference

1. https://leetcode.com/problems/slowest-key/