844. Backspace String Compare

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 = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".

 1 class Solution {
 2     public boolean backspaceCompare(String S, String T) {
 3         String s1 = processRawString(S);
 4         String t1 = processRawString(T);
 5         return s1.equals(t1);
 6     }
 7     
 8     private String processRawString(String rawString) {
 9         if (rawString == null || rawString.isEmpty()) {
10             return "";
11         }
12         Stack<Character> processedString = new Stack<>();
13         for (int i = 0; i < rawString.length(); i++) {
14             char ch = rawString.charAt(i);
15             if (ch == '#') {
16                 if (!processedString.isEmpty()) {
17                     processedString.pop();
18                 }
19             } else {
20                 processedString.push(ch);
21             }
22         }
23         StringBuilder result = new StringBuilder();
24         while (!processedString.isEmpty()) {
25             result.insert(0, processedString.pop());
26         }
27         return result.toString();
28     }
29 }

如果要求不能使用extra space, 那么我们可以从后往前process.

 1 class Solution {
 2     public boolean backspaceCompare(String S, String T) {
 3         int i = S.length() - 1, j = T.length() - 1;
 4 
 5         while (i >= 0 || j >= 0) {
 6             int c1 = 0;
 7             while (i >= 0 && (c1 > 0 || S.charAt(i) == '#')) {
 8                 if (S.charAt(i) == '#') {
 9                     c1++;
10                 } else {
11                     c1--;
12                 }
13                 i--;
14             }
15 
16             int c2 = 0;
17             while (j >= 0 && (c2 > 0 || T.charAt(j) == '#')) {
18                 if (T.charAt(j) == '#') {
19                     c2++;
20                 } else {
21                     c2--;
22                 }
23                 j--;
24             }
25 
26             if (i >= 0 && j >= 0) {
27                 if (S.charAt(i) != T.charAt(j)) {
28                     return false;
29                 } else {
30                     i--;
31                     j--;
32                 }
33             } else {
34                 if (i >= 0 || j >= 0) {
35                     return false;
36                 }
37             }
38         }
39 
40         return i < 0 && j < 0;
41     }
42 }

 

posted @ 2020-12-16 07:55  北叶青藤  阅读(115)  评论(0)  编辑  收藏  举报