# [844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/)

## 题目

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?

## 题目大意


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。

## 解题思路

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。













## 代码

```go

package leetcode

func backspaceCompare(S string, T string) bool {
	s := make([]rune, 0)
	for _, c := range S {
		if c == '#' {
			if len(s) > 0 {
				s = s[:len(s)-1]
			}
		} else {
			s = append(s, c)
		}
	}
	s2 := make([]rune, 0)
	for _, c := range T {
		if c == '#' {
			if len(s2) > 0 {
				s2 = s2[:len(s2)-1]
			}
		} else {
			s2 = append(s2, c)
		}
	}
	return string(s) == string(s2)
}

```