LeetCode 844. Backspace String Compare

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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 }

 

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