844. Backspace String Compare

最新推荐文章于 2023-04-09 13:59:59 发布

BigFatSheep

最新推荐文章于 2023-04-09 13:59:59 发布

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分类专栏： Leetcode Easy

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本文介绍了一种在O(N)时间复杂度和O(1)空间复杂度下比较两个带有#字符作为退格操作的字符串是否相等的算法。通过从字符串尾部开始迭代，有效地处理了退格操作，实现了高效字符串比较。

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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?

Problem URL

Solution

给两个string，#表示删除，问两个string输入之后是否是一样的。

We iterate the two strings from tail to head. Using a for loop to delete ‘#’ and the characters it deleted. Then judge whether the pointers are approached end.

Code

class Solution {
    public boolean backspaceCompare(String S, String T) {
        int i = S.length() - 1;
        int j = T.length() - 1;
        while(true){
            for (int count = 0; i >= 0 && (count > 0 || S.charAt(i) == '#'); i--){
                count += S.charAt(i) == '#' ? 1 : -1;
            }
            for (int count = 0; j >= 0 && (count > 0 || T.charAt(j) == '#'); j--){
                count += T.charAt(j) == '#' ? 1 : -1;
            }
            if (i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)){
                i--;
                j--;
            }
            else{
                return i == -1 && j == -1;
            }
        }
    }
}