1629. 按键持续时间最长的键

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题目

A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

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.

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].

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.

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.

Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"

Output: "c"

Explanation: The keypresses were as follows:

Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).

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).

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).

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).

The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.

'c' is lexicographically larger than 'b', so the answer is 'c'.

Example 2:

Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"

Output: "a"

Explanation: The keypresses were as follows:

Keypress for 's' had a duration of 12.

Keypress for 'p' had a duration of 23 - 12 = 11.

Keypress for 'u' had a duration of 36 - 23 = 13.

Keypress for 'd' had a duration of 46 - 36 = 10.

Keypress for 'a' had a duration of 62 - 46 = 16.

The longest of these was the keypress for 'a' with duration 16.

Constraints:

• releaseTimes.length == n
• keysPressed.length == n
• 2 <= n <= 1000
• 1 <= releaseTimes[i] <= 10^9
• releaseTimes[i] < releaseTimes[i+1]
• keysPressed contains only lowercase English letters.

题目大意

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。

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] 。

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。

请返回单次按键 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。

示例 1：

输入： releaseTimes = [9,29,49,50], keysPressed = "cbcd"

输出： "c"

解释： 按键顺序和持续时间如下：

按下 'c' ，持续时间 9（时间 0 按下，时间 9 松开）

按下 'b' ，持续时间 29 - 9 = 20（松开上一个键的时间 9 按下，时间 29 松开）

按下 'c' ，持续时间 49 - 29 = 20（松开上一个键的时间 29 按下，时间 49 松开）

按下 'd' ，持续时间 50 - 49 = 1（松开上一个键的时间 49 按下，时间 50 松开）

按键持续时间最长的键是 'b' 和 'c'（第二次按下时），持续时间都是 20

'c' 按字母顺序排列比 'b' 大，所以答案是 'c'

示例 2：

输入： releaseTimes = [12,23,36,46,62], keysPressed = "spuda"

输出： "a"

解释： 按键顺序和持续时间如下：

按下 's' ，持续时间 12

按下 'p' ，持续时间 23 - 12 = 11

按下 'u' ，持续时间 36 - 23 = 13

按下 'd' ，持续时间 46 - 36 = 10

按下 'a' ，持续时间 62 - 46 = 16

按键持续时间最长的键是 'a' ，持续时间 16

提示：

• releaseTimes.length == n
• keysPressed.length == n
• 2 <= n <= 1000
• 1 <= releaseTimes[i] <= 10^9
• releaseTimes[i] < releaseTimes[i+1]
• keysPressed 仅由小写英文字母组成

解题思路

1. 初始化：

   • 定义变量 longestDuration 来记录最长持续时间，初始化为第一个按键的持续时间 releaseTimes[0]。
   • 定义变量 res 来记录当前最长持续时间对应的按键，初始化为 keysPressed[0]。

2. 遍历按键记录：

   • 从第二个按键开始，计算每个按键的持续时间： duration = releaseTimes[i] - releaseTimes[i - 1]
   • 比较当前按键持续时间 duration 和记录的 longestDuration：
     • 如果 duration > longestDuration，更新 longestDuration 和 res；
     • 如果 duration == longestDuration，按键字典序 keysPressed[i] 更大时更新 res。

3. 遍历完成后，res 即为答案。

复杂度分析

1. 时间复杂度：O(n)，遍历 releaseTimes 一次。
2. 空间复杂度：O(1)，使用常量空间记录结果和最长持续时间。

代码

/**
 * @param {number[]} releaseTimes
 * @param {string} keysPressed
 * @return {character}
 */
var slowestKey = function (releaseTimes, keysPressed) {
	let longestDuration = releaseTimes[0]; // 第一个按键的持续时间
	let res = keysPressed[0]; // 第一个按键字符

	for (let i = 1; i < releaseTimes.length; i++) {
		const duration = releaseTimes[i] - releaseTimes[i - 1]; // 当前按键持续时间

		// 比较持续时间，若相等则比较字典序
		if (
			duration > longestDuration ||
			(duration == longestDuration && keysPressed[i] > res)
		) {
			longestDuration = duration;
			res = keysPressed[i];
		}
	}

	return res;
};