The Backspace String Compare Algorithm

  • 时间：2020-09-30 16:23:25
  • 分类：网络文摘
  • 阅读：148 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：
五帝本纪赞原文及翻译　 对楚王问原文及翻译　 卜居原文及翻译　 谏逐客书原文及翻译　 乐毅报燕王书原文及翻译　 唐雎不辱使命原文及翻译　 唐雎说信陵君原文及翻译　 鲁共公择言原文及翻译　 鲁仲连义不帝秦原文及翻译　 触龙说赵太后原文及翻译　
评论列表
添加评论