LeetCode-Backspace String Compare

最新推荐文章于 2020-09-07 10:43:40 发布

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本文介绍了一种在存在特殊字符'#'，表示退格操作的情况下，比较两个字符串是否相等的算法。通过使用栈和StringBuilder的数据结构，该算法能够在O(N)的时间复杂度内完成任务。

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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?

题意：判断两个字符串是否相同；字符串中包含符号’#’，表示前一个字符为空字符，例如对于字符串"ab#"=>“a”；

解法：我们主要需要解决的就是如果处理空字符；很容易想到可以利用栈来实现，每当遇到字符’#'时，将栈顶的元素出栈，否则将此字符入栈；因为最后还要返回一个新的字符串，所以我们使用StringBuilder来模拟栈；

Java

class Solution {
    public boolean backspaceCompare(String S, String T) {
        return removeBackspace(S).equals(removeBackspace(T));
    }
    private String removeBackspace(String s) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '#') {
                if (sb.length() == 0) continue;
                sb.deleteCharAt(sb.length() - 1);
            } else {
                sb.append(s.charAt(i));
            }
        }
        return sb.toString();
    }
}