[LeetCode] 844. Backspace String Compare

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 and t only contain lowercase letters and '#' characters. 

Follow up: Can you solve it in O(n) time and O(1) space?

比较含退格的字符串。

给定 s 和 t 两个字符串，当它们分别被输入到空白的文本编辑器后，如果两者相等，返回 true 。# 代表退格字符。

注意：如果对空文本输入退格字符，文本继续为空。

题意是给两个字符串，中间包含'#'，井字的意思是需要退格，请判断在处理完井字之后，两个字符串是否相等。

这个题比较直观的思路是用 stack 做。followup 是会问如何做到不用额外空间，做法是反向遍历，从字符串的尾部遍历到头部。

stack实现，做法应该很直观了，我直接给代码，这里 stack 我用 Deque 实现。

时间O(n)

空间O(n)

Java实现

class Solution {
    public boolean backspaceCompare(String s, String t) {
        Deque<Character> queue1 = new ArrayDeque<>();
        Deque<Character> queue2 = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == '#') {
                queue1.pollLast();
            } else {
                queue1.offer(c);
            }
        }

        for (char c : t.toCharArray()) {
            if (c == '#') {
                queue2.pollLast();
            } else {
                queue2.offer(c);
            }
        }

        StringBuilder sb1 = new StringBuilder();
        while (!queue1.isEmpty()) {
            sb1.append(queue1.poll());
        }
        StringBuilder sb2 = new StringBuilder();
        while (!queue2.isEmpty()) {
            sb2.append(queue2.poll());
        }
        return sb1.toString().equals(sb2.toString());
    }
}

 

JavaScript实现

/**
 * @param {string} S
 * @param {string} T
 * @return {boolean}
 */
var backspaceCompare = function(S, T) {
    return helper(S) == helper(T);
};

var helper = function(input) {
    let stack = [];
    for (let c of input) {
        if (c !== '#') {
            stack.push(c);
        } else if (c == '#' && stack.length > 0) {
            stack.pop();
        }
    }
    return stack.join('');
};

 

反向扫描

反向扫描的思路是，从字符串的尾部遍历到头部，记录一个变量 skip，当遇到 # 的时候，skip++，同时 continue 略过当前这个 #；之后再看 skip，如果 skip > 0，则说明需要跳过当前这个字符，就再 continue。如果 skip == 0 则直接 append 当前字符。

时间O(n)

空间O(1)

Java实现

class Solution {
    public boolean backspaceCompare(String s, String t) {
        StringBuilder sb1 = new StringBuilder();
        StringBuilder sb2 = new StringBuilder();
        int sskip = 0;
        int tskip = 0;
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (c == '#') {
                sskip++;
                continue;
            }
            if (sskip > 0) {
                sskip--;
                continue;
            } else {
                sb1.append(c);
            }
        }

        for (int i = t.length() - 1; i >= 0; i--) {
            char c = t.charAt(i);
            if (c == '#') {
                tskip++;
                continue;
            }
            if (tskip > 0) {
                tskip--;
                continue;
            } else {
                sb2.append(c);
            }
        }
        return sb1.toString().equals(sb2.toString());
    }
}

 

JavaScript实现

/**
 * @param {string} S
 * @param {string} T
 * @return {boolean}
 */
var backspaceCompare = function (S, T) {
    let sskip = 0;
    let tskip = 0;
    let sb1 = [];
    let sb2 = [];
    for (let i = S.length - 1; i >= 0; i--) {
        let c = S.charAt(i);
        if (c === '#') {
            sskip++;
            continue;
        }
        if (sskip > 0) {
            sskip--;
            continue;
        } else {
            sb1.push(c);
        }
    }

    for (let j = T.length - 1; j >= 0; j--) {
        let v = T.charAt(j);
        if (v === '#') {
            tskip++;
            continue;
        }
        if (tskip > 0) {
            tskip--;
            continue;
        } else {
            sb2.push(v);
        }
    }
    return sb1.toString() === sb2.toString();
};

 

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