2025-01-12 2025-01-12 problem 几秒读完 (大约108个字) 0次访问LC3019. 按键变更的次数题目链接:3019. 按键变更的次数 - 力扣(LeetCode) 题解 按题意模拟,看其他人题解有用位运算去写更加简洁。 参考代码 1234567891011121314151617## @lc app=leetcode.cn id=3019 lang=python3## [3019] 按键变更的次数## @lc code=startclass Solution: def countKeyChanges(self, s: str) -> int: ans = 0 s = s.lower() for i in range(1, len(s)): if s[i] != s[i-1]: ans += 1 return ans# @lc code=end LC3019. 按键变更的次数https://blog.xiang578.com/problem/lc3019.html作者Ryen Xiang发布于2025-01-12更新于2025-01-12许可协议 LeetCode, implementation, Problems/2025
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