The Backspace String Compare Algorithm

  • 时间：2020-09-30 16:23:25
  • 分类：网络文摘
  • 阅读：115 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：
春的脚步作文400字　 好习惯让人受益一生　 我最喜欢去的一个地方　 国庆假期作文600字　 书籍触动我的心灵　 有趣的数学问题　 这个学校春游的学生有多少人　 它们将于几时几分在途中再次相遇　 AB两地的路程是550千米　 李叔叔从家骑车去横溪办事　
评论列表
添加评论