Question

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.

Note:

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

Difficulty:Easy

Category:Two-Points, String

Analyze

这道题目使用两个指针分别定位name和typed，当左边和右边不相等的时候，判断是否和上一位相等，如果和name的上一位相等，那么指针l就向右移动一位。

Solution

// Solution: Two Points
// Runtimes: 4ms
class Solution {
 public:
  bool isLongPressedName(string name, string typed) {
    int l = 0, r = 0;
    char prev;
    while (l < name.length()) {
      if (typed[r] == name[l]) {
        prev = typed[r];
        r++;
        l++;
      } else if (typed[r] == prev) {
        r++;
      } else
        return false;
    }
    return true;
  }
};