LeetCode 1087. Brace Expansion

标签：input   one   man   leetcode   nts   character   oar   continue   limit   

原题链接在这里：https://leetcode.com/problems/brace-expansion/

题目：

A string S represents a list of words.

Each letter in the word has 1 or more options.  If there is one option, the letter is represented as is.  If there is more than one option, then curly braces delimit the options.  For example, "{a,b,c}" represents options ["a", "b", "c"].

For example, "{a,b,c}d{e,f}" represents the list ["ade", "adf", "bde", "bdf", "cde", "cdf"].

Return all words that can be formed in this manner, in lexicographical order.

Example 1:

Input: "{a,b}c{d,e}f"
Output: ["acdf","acef","bcdf","bcef"]

Example 2:

Input: "abcd"
Output: ["abcd"]

Note:

1. 1 <= S.length <= 50
2. There are no nested curly brackets.
3. All characters inside a pair of consecutive opening and ending curly brackets are different.

题解：

If there is curly braces, all the chars in it could be candidate.

Starting from index 0. Do DFS, DFS state needs orginal string, current index, current StringBuilder and res collection.

If current index i points to ‘{‘, then find next index j points to ‘}‘, for each of candidate inside brances, append it to StringBuilder and continue DFS at index j+1. After DFS, do backtracking.

If current index i points to char, append it to StringBuilder and continue DFS at index i+1. After DFS, do bracktracking.

When current index points to the end of string, add copy of StringBuilder to res collection.

Time Complexity: exponential.

Space: O(n). n = S.length(). stack space.

AC Java:

 1 class Solution {
 2     public String[] expand(String S) {
 3         if(S == null || S.length() == 0){
 4             return new String[0];
 5         }
 6         
 7         List<String> res = new ArrayList<String>();
 8         dfs(S, 0, new StringBuilder(), res);
 9         Collections.sort(res);
10         return res.toArray(new String[0]);
11     }
12     
13     private void dfs(String s, int i, StringBuilder sb, List<String> res){
14         if(i >= s.length()){
15             res.add(sb.toString());
16             return;
17         }
18         
19         
20         if(s.charAt(i) == ‘{‘){
21             int j = i+1;
22             while(j<s.length() && s.charAt(j)!=‘}‘){
23                 j++;
24             }
25             
26             String [] candidates = s.substring(i+1, j).split(",");
27             for(String candidate : candidates){
28                 sb.append(candidate);
29                 dfs(s, j+1, sb, res);
30                 sb.deleteCharAt(sb.length()-1);
31             }
32         }else{
33             sb.append(s.charAt(i));
34             dfs(s, i+1, sb, res);
35             sb.deleteCharAt(sb.length()-1);
36         }
37     }
38 }

LeetCode 1087. Brace Expansion

标签：input   one   man   leetcode   nts   character   oar   continue   limit   

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11791503.html