844. 比较含退格的字符串
844. 比较含退格的字符串
🟢 🔖 栈
双指针
字符串
模拟
🔗 力扣
LeetCode
题目
Given two strings s
and t
, return true
if they are equal when both are typed into empty text editors. '#'
means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
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 = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, t.length <= 200
s
andt
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,注意 JS 中删除字符串的最后一个字符可以用str.slice(0, -1)
; - 不是
#
号就入栈一个字符; - 最后比较 2 个字符串即可;
代码
/**
* @param {string} s
* @param {string} t
* @return {boolean}
*/
var backspaceCompare = function (s, t) {
return backspace(s) === backspace(t);
};
var backspace = function (str) {
let res = '';
for (item of str) {
if (item === '#') res = res.slice(0, -1);
else res += item;
}
return res;
};
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