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83 lines
4.2 KiB
Markdown
83 lines
4.2 KiB
Markdown
# [1629. Slowest Key](https://leetcode.com/problems/slowest-key/)
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## 题目
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A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
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You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was released. Both arrays are **0-indexed**. The `0th` key was pressed at the time `0`, and every subsequent key was pressed at the **exact** time the previous key was released.
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The tester wants to know the key of the keypress that had the **longest duration**. The `ith` keypress had a **duration** of `releaseTimes[i] - releaseTimes[i - 1]`, and the `0th` keypress had a duration of `releaseTimes[0]`.
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Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key **may not** have had the same **duration**.
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*Return the key of the keypress that had the **longest duration**. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.*
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**Example 1:**
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```
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Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
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Output: "c"
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Explanation: The keypresses were as follows:
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Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
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Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
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Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
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Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
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The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
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'c' is lexicographically larger than 'b', so the answer is 'c'.
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```
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**Example 2:**
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```
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Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
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Output: "a"
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Explanation: The keypresses were as follows:
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Keypress for 's' had a duration of 12.
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Keypress for 'p' had a duration of 23 - 12 = 11.
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Keypress for 'u' had a duration of 36 - 23 = 13.
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Keypress for 'd' had a duration of 46 - 36 = 10.
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Keypress for 'a' had a duration of 62 - 46 = 16.
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The longest of these was the keypress for 'a' with duration 16.
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```
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**Constraints:**
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- `releaseTimes.length == n`
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- `keysPressed.length == n`
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- `2 <= n <= 1000`
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- `1 <= releaseTimes[i] <= 109`
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- `releaseTimes[i] < releaseTimes[i+1]`
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- `keysPressed` contains only lowercase English letters.
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## 题目大意
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LeetCode 设计了一款新式键盘，正在测试其可用性。测试人员将会点击一系列键（总计 n 个），每次一个。
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给你一个长度为 n 的字符串 keysPressed ，其中 keysPressed[i] 表示测试序列中第 i 个被按下的键。releaseTimes 是一个升序排列的列表，其中 releaseTimes[i] 表示松开第 i 个键的时间。字符串和数组的 下标都从 0 开始 。第 0 个键在时间为 0 时被按下，接下来每个键都 恰好 在前一个键松开时被按下。测试人员想要找出按键 持续时间最长 的键。第 i 次按键的持续时间为 releaseTimes[i] - releaseTimes[i - 1] ，第 0 次按键的持续时间为 releaseTimes[0] 。
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注意，测试期间，同一个键可以在不同时刻被多次按下，而每次的持续时间都可能不同。请返回按键 持续时间最长 的键，如果有多个这样的键，则返回 按字母顺序排列最大 的那个键。
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## 解题思路
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- 题干很长，不过还是简单题。循环扫描一遍数组，计算出每个按键的持续时间。动态更新按键最长时间的键。如果持续时间最长的有多个键，需要返回字母序最大的那一个键。
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## 代码
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```go
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package leetcode
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func slowestKey(releaseTimes []int, keysPressed string) byte {
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longestDuration, key := releaseTimes[0], keysPressed[0]
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for i := 1; i < len(releaseTimes); i++ {
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duration := releaseTimes[i] - releaseTimes[i-1]
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if duration > longestDuration {
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longestDuration = duration
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key = keysPressed[i]
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} else if duration == longestDuration && keysPressed[i] > key {
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key = keysPressed[i]
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}
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}
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return key
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}
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``` |