The Backspace String Compare Algorithm

  • 时间：2020-09-30 16:23:25
  • 分类：网络文摘
  • 阅读：133 次

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. Can you solve it in O(N) time and O(1) space?

Javascript algorithm to perform backspace string comparison

To simulate the backspace key, we can use a stack, and perform a pop operation when we want to delete previous character. In Javascript, we can use Array.prototype.pop() to remove the last element (which can be called on empty array and that returns undefined). When we iterate all characters, we need to join the stack/array as a string. Then, the last step is to perform a string comparisons.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

/**
 * @param {string} S
 * @param {string} T
 * @return {boolean}
 */
var backspaceCompare = function(S, T) {
    const build = (S) => {
        let st = [];        
        for (let i = 0, len = S.length; i < len; ++ i) {
            if (S[i] == '#') {
                st.pop(); // if called on empty array, return undefined.
            } else {
                st.push(S[i]);
            }
        }
        return st.join('');
    }
    return build(S) === build(T);
};
/**
 * @param {string} S
 * @param {string} T
 * @return {boolean}
 */
var backspaceCompare = function(S, T) {
    const build = (S) => {
        let st = [];        
        for (let i = 0, len = S.length; i < len; ++ i) {
            if (S[i] == '#') {
                st.pop(); // if called on empty array, return undefined.
            } else {
                st.push(S[i]);
            }
        }
        return st.join('');
    }
    return build(S) === build(T);
};

String backspace comparison in C++

Slightly differently, we do not construct/join the characters in the stack, instead, we can perform comparisons on both stacks generated from two input strings. 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31

class Solution {
public:
    bool backspaceCompare(string S, string T) {
        stack<char> st1 = realString(S);
        stack<char> st2 = realString(T);
        if (st1.size() != st2.size()) return false;
        while (st1.size() > 0) {
            char p1 = st1.top();
            char p2 = st2.top();
            st1.pop();
            st2.pop();
            if (p1 != p2) return false;
        }
        return true;
    }
    
private:
    stack<char> realString(string S) {
        stack<char> st;    
        for (int i = 0; i < S.size(); ++ i) {
            if (S[i] == '#') {
                if (st.size() > 0) {
                    st.pop();
                }
            } else {
                st.push(S[i]);
            }
        }
        return st;
    }
};
class Solution {
public:
    bool backspaceCompare(string S, string T) {
        stack<char> st1 = realString(S);
        stack<char> st2 = realString(T);
        if (st1.size() != st2.size()) return false;
        while (st1.size() > 0) {
            char p1 = st1.top();
            char p2 = st2.top();
            st1.pop();
            st2.pop();
            if (p1 != p2) return false;
        }
        return true;
    }
    
private:
    stack<char> realString(string S) {
        stack<char> st;    
        for (int i = 0; i < S.size(); ++ i) {
            if (S[i] == '#') {
                if (st.size() > 0) {
                    st.pop();
                }
            } else {
                st.push(S[i]);
            }
        }
        return st;
    }
};

Java implementation of string backspace comparisons

We can use String.valueOf(stack) to convert the stack of characters into String.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

class Solution {
    public boolean backspaceCompare(String S, String T) {
        return build(S).equals(build(T));
    }
 
    public String build(String S) {
        Stack<Character> ans = new Stack();
        for (char c: S.toCharArray()) {
            if (c != '#')
                ans.push(c);
            else if (!ans.empty())
                ans.pop();
        }
        return String.valueOf(ans);
    }
}
class Solution {
    public boolean backspaceCompare(String S, String T) {
        return build(S).equals(build(T));
    }

    public String build(String S) {
        Stack<Character> ans = new Stack();
        for (char c: S.toCharArray()) {
            if (c != '#')
                ans.push(c);
            else if (!ans.empty())
                ans.pop();
        }
        return String.valueOf(ans);
    }
}

All above implementations run at complexity O(M + N) in both time and space where M and N are the lengths of input strings S and T respectively.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
扇形和半圆有重叠部分　 为wordpress新建一个登录页面　 wordpress自动显示图片exif信息插件-Display Exif　 为wordpress设置一个简单的后台登陆地址　 免插件自动生成wordpress文章二维码图片　 wordpress主题更新style.css样式表文件后不生效的解决方法　 如何在WordPress文章末尾自动添加作者简介　 如何让WordPress分类归档排除子分类文章　 如何让WordPress显示即将发布的文章列表　 wordpress头像缓存插件：nix-gravatar-cache 修改版　
评论列表
添加评论