leetcode 925. Long Pressed Name 长按键入-耗时100

Problem: 925. Long Pressed Name 长按键入

耗时100%，双指针，若左右字符相等l++; r++，若r > 0 && typed[r-1]==typed[r]则r++，否则false，最后若while(r < n2 && typed[r-1]==typed[r]) r++;，if(l!=n1 || r!=n2) return false;

Code

class Solution { public: bool isLongPressedName(string name, string typed) { int l = 0, r = 0, n1 = name.size(), n2 = typed.size(); while(l < n1 && r < n2) { if(name[l] == typed[r]) { l++; r++; } else if(r > 0 && typed[r-1]==typed[r]) { r++; } else { return false; } } while(r < n2 && typed[r-1]==typed[r]) r++; if(l!=n1 || r!=n2) return false; return true; } };