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111 lines
1.6 KiB
Markdown
111 lines
1.6 KiB
Markdown
# [844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/)
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## 题目
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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.
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**Example 1**:
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```
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Input: S = "ab#c", T = "ad#c"
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Output: true
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Explanation: Both S and T become "ac".
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```
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**Example 2**:
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```
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Input: S = "ab##", T = "c#d#"
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Output: true
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Explanation: Both S and T become "".
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```
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**Example 3**:
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```
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Input: S = "a##c", T = "#a#c"
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Output: true
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Explanation: Both S and T become "c".
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```
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**Example 4**:
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```
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Input: S = "a#c", T = "b"
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Output: false
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Explanation: S becomes "c" while T becomes "b".
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```
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**Note**:
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- 1 <= S.length <= 200
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- 1 <= T.length <= 200
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- S and T only contain lowercase letters and '#' characters.
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**Follow up**:
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- Can you solve it in O(N) time and O(1) space?
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## 题目大意
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给 2 个字符串，如果遇到 # 号字符，就回退一个字符。问最终的 2 个字符串是否完全一致。
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## 解题思路
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这一题可以用栈的思想来模拟，遇到 # 字符就回退一个字符。不是 # 号就入栈一个字符。比较最终 2 个字符串即可。
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## 代码
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```go
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package leetcode
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func backspaceCompare(S string, T string) bool {
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s := make([]rune, 0)
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for _, c := range S {
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if c == '#' {
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if len(s) > 0 {
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s = s[:len(s)-1]
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}
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} else {
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s = append(s, c)
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}
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}
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s2 := make([]rune, 0)
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for _, c := range T {
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if c == '#' {
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if len(s2) > 0 {
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s2 = s2[:len(s2)-1]
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}
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} else {
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s2 = append(s2, c)
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}
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}
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return string(s) == string(s2)
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}
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``` |