leetcode925 Long Pressed Name
給定字串 A 和 B ,輸入 A 時某些字母會手抖打多遍,問 B 是否可能是 A 手抖後的結果。
思路:暴力即可,兩個指標,滿足不了條件就return false
class Solution { public: bool isLongPressedName(string name, string typed) { int i=0,j=0; int len1=name.length(),len2=typed.length(); if(len1>len2) return false; while(i<len1||j<len2) { if(name[i]==typed[j]){ i++,j++; } else if(name[i-1]==typed[j]) { j++; } else { return false; } } return true; } };
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給定字串 A 和 B ,輸入 A 時某些字母會手抖打多遍,問 B 是否可能是 A 手抖後的結果。 思路:暴力即可,兩個指標,滿足不了條件就return false class Solution
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