844.Backspace String Compare

最新推荐文章于 2023-04-09 13:59:59 发布

Zhou_xinke

最新推荐文章于 2023-04-09 13:59:59 发布

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分类专栏： leetcode c++ stack

本文链接：https://blog.youkuaiyun.com/Zhou_xinke/article/details/83308216

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本文介绍了一种高效的字符串比较算法，用于处理包含退格字符的字符串。通过使用栈数据结构，算法能在O(N)时间复杂度内完成比较，适用于编程竞赛及实际应用中。文章提供了两段代码实现，一段为标准实现，另一段为优化后的0ms实现。

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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?

==可以直接用来判断两个栈是否相等

class Solution {
public:
    bool backspaceCompare(string S, string T) {
        stack<char> s1;
        stack<char> s2;
        for(int i = 0;i<S.length();i++)
        {
            if(s1.empty()&&S[i] == '#')
                ;
            else if(S[i] == '#')
                s1.pop();
            else
                s1.push(S[i]);
        }
        
        for(int i = 0;i<T.length();i++)
        {
            if(s2.empty()&&T[i] == '#')
                ;
            else if(T[i] == '#')
                s2.pop();
            else
                s2.push(T[i]);
        }
        if(s1 == s2)
            return true;
        else
            return false;
    }
};

再贴一个大佬写的代码，0ms，超过了100%，惊呆了！！

class Solution {
public:
    bool backspaceCompare(string S, string T) {
        stack<int>s,t;
        for(auto c : S)
        {
            if(c == '#')
            {
                if(!s.empty())
                    s.pop();
            }
            else
                s.push(c);
        }
        for(auto c : T)
        {
            if(c == '#')
            {
                if(!t.empty())
                    t.pop();
            }
            else
                t.push(c);
        }
        if(s == t)
            return true;
        else
            return false;
    }
};