844. Compare strings with backspace [JavaScript]

Given two strings of S and T, when they are input into a blank text editor, determine whether they are equal, and return the result. # Represents the backspace character.

Note: If you enter a backspace character for empty text, the text will remain empty.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T will become "ac".
Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T will become "".
Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T will become "c".
Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S will become "c", but T will still be "b".

prompt:

1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and the character'#'.

Source: LeetCode
Link: https://leetcode-cn.com/problems/backspace-string-compare

I made this question by accident after reading the idea of ​​an up master at station b. I gave the two statistical names sspace and tspace, and compared them from back to front. When the # number is encountered, it will increase by 1, and vice versa. Subtract 1 and compare the corresponding bits of the two strings when they are both 0, and return false if they are different.
code show as below:

        /**
         * @param {string} S
         * @param {string} T
         * @return {boolean}
         */
        var backspaceCompare = function(S, T) {
    
    
            var i=S.length-1;
            var j=T.length-1;;
            var sspace=0;
            var tspace=0;
            while(i>=0||j>=0){
    
    
            while(i>=0){
    
    
                if(S[i]=='#'){
    
    
                sspace++;
                i--;
            }else if(sspace>0){
    
    
                sspace--;
                i--;
            }else{
    
    
                break;
            }
        }
        while(j>=0){
    
    
            if(T[j]=='#'){
    
    
                tspace++;
                j--;
            }else if(tspace>0){
    
    
                tspace--;
                j--;
            }else{
    
    
                break;
            }
        }
        if(S[i]!==T[j]){
    
    
            return false;
        }
        if(i<0&&j>=0||i>=0&&j<0){
    
    
            return false;
        }
        i--;
        j--;
            }
            return true;
            };

After reading the solutions of others, I think it is indeed easier to use regular expressions. For example, the code of this old man, for your reference:

/**
 * @param {string} S
 * @param {string} T
 * @return {boolean}
 */
var backspaceCompare = function(S, T) {
    
    
    let reg = /[a-z](?=\#)\#/g;
    S = S.replace(/^\#/,'').replace(reg, '');
    T = T.replace(/^\#/,'').replace(reg, '');
    if(!S.includes('#') && !T.includes('#')){
    
    
        return S == T;
    }
    return backspaceCompare(S, T);
};

作者：caoyq0521
链接：https://leetcode-cn.com/problems/backspace-string-compare/solution/bi-jiao-han-tui-ge-de-zi-fu-chuan-by-caoyq0521/
来源：力扣（LeetCode）
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