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1629. 按键持续时间最长的键
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1629. 按键持续时间最长的键

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。

遍历即可。

class Solution:
def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
n = len(releaseTimes)
if n == 1:
return keysPressed
longest = releaseTimes[0]
ans = keysPressed[0]
for i in range(n-1):
if longest < releaseTimes[i+1] - releaseTimes[i]:
ans = keysPressed[i+1]
longest = releaseTimes[i+1] - releaseTimes[i]
elif longest == releaseTimes[i+1] - releaseTimes[i] and ord(ans) < ord(keysPressed[i+1]):
ans = keysPressed[i+1]
return ans
class Solution: def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str: n = len(releaseTimes) if n == 1: return keysPressed longest = releaseTimes[0] ans = keysPressed[0] for i in range(n-1): if longest < releaseTimes[i+1] - releaseTimes[i]: ans = keysPressed[i+1] longest = releaseTimes[i+1] - releaseTimes[i] elif longest == releaseTimes[i+1] - releaseTimes[i] and ord(ans) < ord(keysPressed[i+1]): ans = keysPressed[i+1] return ans
class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        n = len(releaseTimes)
        if n == 1:
            return keysPressed
        longest = releaseTimes[0]
        ans = keysPressed[0]
        for i in range(n-1):
            if longest < releaseTimes[i+1] - releaseTimes[i]:
                ans = keysPressed[i+1]
                longest = releaseTimes[i+1] - releaseTimes[i]
            elif longest == releaseTimes[i+1] - releaseTimes[i] and ord(ans) < ord(keysPressed[i+1]):
                ans = keysPressed[i+1]
        return ans
                

官方标答:

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

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/slowest-key
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

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