Backspace String Compare
Difficulty: EasyTopic: Two PointersStringStackSimulationLeetcode:874.Backspace String CompareGiven 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 <= 200sandtonly contain lowercase letters and'#'characters.
Follow up: Can you solve it in O(n) time and O(1) space?
Solution
write solution in here
/**
* @param {string} s
* @param {string} t
* @return {boolean}
*/
var backspaceCompare = function(s, t) {
let ss = [], ts = []
for(let i = 0; i < s.length; i ++) {
s[i] !== "#" ? ss.push(s[i]) : ss.pop()
}
for(let i = 0; i < t.length; i ++) {
t[i] !== "#" ? ts.push(t[i]) : ts.pop()
}
if (ss.length !== ts.length) return false
for(let i = 0; i < ss.length; i ++) {
if (ss[i] !== ts[i]) return false
}
return true
};