LeetCode 844. Backspace String Compare

Posted 2021-03-19 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/backspace-string-compare/

题目：

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?

题解：

Could easily do it with stack. The follow up is to use O(N) time and O(1) space.

We could iterate the string back to front. Accumlate he # and use it when it is no #.

Then check if two chars are the same, if yes, move both pointers.

If not, check if two pointers alreay come to -1.

Time Complexity: O(m+n). m = S.length(). n = T.length().

Space: O(1).

AC Java:

 1 class Solution {
 2     public boolean backspaceCompare(String S, String T) {
 3         if(S == null || T == null){
 4             return S == T;
 5         }
 6         
 7         int m = S.length();
 8         int n = T.length();
 9         int i = m - 1;
10         int j = n - 1;
11         int cntS = 0;
12         int cntT = 0;
13         
14         while(i >= 0 || j >= 0){
15             while(i >= 0 && (S.charAt(i) == ‘#‘ || cntS > 0)){
16                 if(S.charAt(i) == ‘#‘){
17                     cntS++;
18                 }else{
19                     cntS--;
20                 }
21                 
22                 i--;
23             }
24             
25             while(j >= 0 && (T.charAt(j) == ‘#‘ || cntT > 0)){
26                 if(T.charAt(j) == ‘#‘){
27                     cntT++;
28                 }else{
29                     cntT--;
30                 }
31                 
32                 j--;
33             }
34             
35             if(i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)){
36                 i--;
37                 j--;
38             }else{
39                 return i == -1 && j == -1;
40             }
41         }
42         
43         return true;
44     }
45 }