[leetcode] 844. Backspace String Compare

原创

是念 2022-08-11 17:54:38 博主文章分类：leetcode ©著作权

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Description

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".

Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".

Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".

Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".

Note:

1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and ‘#’ characters.

Follow up:
Can you solve it in O(N) time and O(1) space?

分析

题目的意思为：判断两个字符串最终的输出是否是一样的，其中#代表键盘回退/删除

双指针法,从后往前遍历，代码写了三个循环，很容易懂

代码

class Solution {
public:
    bool backspaceCompare(string S, string T) {
        int i=S.size()-1;
        int j=T.size()-1;
        int skipS=0;
        int skipT=0;
        while(i>=0||j>=0){
            while(i>=0){
                if(S[i]=='#'){
                    skipS++;
                    i--;
                }else if(skipS>0){
                    skipS--;i--;
                }else break;
            }
            while(j>=0){
                if(T[j]=='#'){
                    skipT++;
                    j--;
                }else if(skipT>0){
                    skipT--;
                    j--;
                }else break;
            }
            if(i>=0&&j>=0&&S[i]!=T[j]){
                return false;
            }
            if((i>=0)!=(j>=0)){
                return false;
            }
            i--;
            j--;
        }
        return true;
    }
};