LeetCode C++ 925. Long Pressed Name【StringTwo Pointers】简单

z Your friend is typing his name into a keyboard. Sometimes, when typing a character c , the key might get long pressed, and the character will be typed 1 or more times.

You examine the typed characters of the keyboard. Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.

Example 1:

Input: name = "alex", typed = "aaleex" Output: true Explanation: 'a' and 'e' in 'alex' were long pressed.

Example 2:

Input: name = "saeed", typed = "ssaaedd" Output: false Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.

Example 3:

Input: name = "leelee", typed = "lleeelee" Output: true

Example 4:

Input: name = "laiden", typed = "laiden" Output: true Explanation: It's not necessary to long press any character.

Constraints:

1 <= name.length <= 10001 <= typed.length <= 1000The characters of name and typed are lowercase letters.

题意：朋友正在使用键盘输入他的名字 name 。偶尔在键入字符时，按键可能会被长按，因此字符可能被输入 1 次或多次。本题要求检查键盘输入的字符 typed ，如果它对应的可能是朋友的名字（其中一些字符可能被长按），那么就返回 True 。

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解法 双指针

一个小的剪枝是：如果 name 的长度大于 typed 的长度时，直接返回 false 。之后维护两个指针，比较指针指向的两个字符，如果不相等，则返回 false ；如果相等，就分别计算此处这个字符连续出现的长度，如果 name 中连续出现的长度大于 typed 中连续出现的长度，则不可能是长按导致的错误，返回 false ，否则继续往后遍历比较。具体代码如下：

class Solution { public: bool isLongPressedName(string name, string typed) { if (name.size() > typed.size()) return false; int pn = 0, pt = 0, n = name.size(), m = typed.size(); while (pn < n && pt < m) { if (name[pn] != typed[pt]) return false; int i = pn + 1, j = pt + 1; while (i < n && name[i] == name[pn]) ++i; while (j < m && typed[j] == typed[pt]) ++j; if (i - pn > j - pt) return false; pn = i; pt = j; } return pn >= n && pt >= m; } };

效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户 内存消耗：6.3 MB, 在所有 C++ 提交中击败了36.09% 的用户