844. Backspace String Compare - Easy

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. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.

Follow up:

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

 

M1: stack

时间：O(M+N)，空间：O(M+N)

class Solution {
     public boolean backspaceCompare(String S, String T) {
         Stack<Character> s = new Stack<>();
         Stack<Character> t = new Stack<>();
        
         for(int i = 0; i < S.length(); i++) {
             if(S.charAt(i) != '#')
                 s.push(S.charAt(i));
             else if(!s.isEmpty())
                 s.pop();
         }
         for(int i = 0; i < T.length(); i++) {
             if(T.charAt(i) != '#')
                 t.push(T.charAt(i));
             else if(!t.isEmpty())
                 t.pop();
         }
        if(s.size() != t.size()) return false;
        for(int i = 0; i < s.size(); i++) {
            if(s.pop() != t.pop())
                return false;
        }
        return true;
     }
}

 

M2: two pointers

从后往前扫描字符串，用常数cnt统计退格数。遇到‘#’，cnt++；遇到字母并且cnt > 0，cnt--并且指针--；遇到字母但是cnt=0，break退出循环，比较两个字符串对应字符是否相等，不等返回false。如果最后扫描完两个字符串，两个指针都=-1，返回true。如果只扫描完一个字符串，返回false。

时间：O(M+N)，空间：O(1)

class Solution {
     public boolean backspaceCompare(String S, String T) {
         int p1 = S.length() - 1, p2 = T.length() - 1;
         int cnt1 = 0, cnt2 = 0;
         while(p1 >= 0 || p2 >= 0) {
             while(p1 >= 0) {
                 if(S.charAt(p1) == '#') {
                     p1--;cnt1++;
                 }
                 else if(S.charAt(p1) != '#' && cnt1 > 0) {
                     p1--;cnt1--;
                 }
                 else
                     break;
             }
             while(p2 >= 0) {
                 if(T.charAt(p2) == '#') {
                     p2--;cnt2++;
                 }
                 else if(T.charAt(p2) != '#' && cnt2 > 0) {
                     p2--;cnt2--;
                 }
                 else
                     break;
             }
             if(p1 == -1 && p2 == -1) return true;
             if(p1 == -1 || p2 == -1) return false;
             if(S.charAt(p1) != T.charAt(p2)) return false;
             p1--;p2--;
         }
         return true;
     }
}

 

转载于:https://www.cnblogs.com/fatttcat/p/10015990.html