LeetCode #844 - Backspace String Compare

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本文探讨了在存在退格字符的情况下，两个字符串的比较算法。通过实例展示了如何处理特殊字符，实现字符串的有效对比。文章深入解析了算法的时间和空间复杂度，并提出了一种O(N)时间O(1)空间的解决方案。

题目描述：

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. 1 <= S.length <= 200

2. 1 <= T.length <= 200

3. S and T only contain lowercase letters and '#' characters.

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

class Solution {
public:
    bool backspaceCompare(string S, string T) {
        int i=0;
        int j=0;
        while(j<S.size())
        {
            if(S[j]=='#')
            {
                if(i>0) i--;
            }
            else
            {
                S[i]=S[j];
                i++;
            }
            j++;
        }
        //S[i]=S[j]时，i++，此时i就是字符串长度
        int length_s=i;
        i=0;
        j=0;
        while(j<T.size())
        {
            if(T[j]=='#')
            {
                if(i>0) i--;
            }
            else
            {
                T[i]=T[j];
                i++;
            }
            j++;
        }
        int length_t=i;
        return S.substr(0,length_s)==T.substr(0,length_t);
    }
};