301. Remove Invalid Parentheses

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Examples:

"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]

首先，遍历所有情况，然后看是否最后得到的string是否满足所有valid parentheses的情况，满足的话，就比较已经存在的valid parentheses. 否则就放弃。
 1 public class Solution {
 2     public List<String> removeInvalidParentheses(String s) {
 3         List<String> list = new ArrayList<>();
 4         if (s == null) return list;
 5         list.add("");
 6         helper(s, 0, "", 0, list);
 7         return list;
 8     }
 9 
10     void helper(String str, int index, String tempString, int leftCount, List<String> list) {
11         if (str.isEmpty() || leftCount > str.length() - index) return;
12 
13         if (list.size() != 0 && list.get(0).length() > tempString.length() + str.length() - index) return;
14 
15         if (index == str.length()) {
16             if (leftCount == 0) {
17                 if (list.get(0).length() < tempString.length()) {
18                     list.clear();
19                     list.add(tempString);
20                 } else if (list.get(0).length() == tempString.length() && !list.contains(tempString)) {
21                     list.add(tempString);
22                 }
23             }
24             return;
25         }
26 
27         if (str.charAt(index) == '(') {
28             helper(str, index + 1, tempString + "(", leftCount + 1, list);
29             helper(str, index + 1, tempString, leftCount, list);
30         } else if (str.charAt(index) == ')') {
31             if (leftCount != 0) {
32                 helper(str, index + 1, tempString + ")", leftCount - 1, list);
33             }
34             helper(str, index + 1, tempString, leftCount, list);
35         } else {
36             helper(str, index + 1, tempString + str.charAt(index), leftCount, list);
37         }
38     }
39 }

 

 1 public class Solution {
 2     public List<String> removeInvalidParentheses(String s) {
 3         List<String> res = new ArrayList<String>();
 4         int[] max = { 0 };
 5         dfs(s, "", 0, res, max);
 6         if (res.size() == 0) {
 7             res.add("");
 8         }
 9         return res;
10     }
11 
12     private void dfs(String str, String subRes, int countLeft, List<String> res, int[] max) {
13         if (countLeft > str.length()) return;
14         if (str.length() == 0) {
15             if (countLeft == 0 && subRes.length() != 0) {
16                 if (subRes.length() > max[0]) {
17                     max[0] = subRes.length(); 
                       res.clear();
18                 }
19                 if (max[0] == subRes.length() && !res.contains(subRes)) {
20                     res.add(subRes.toString());
21                 }
22             }
23             return;
24         }
25 
26         if (str.charAt(0) == '(') {
27             dfs(str.substring(1), subRes + "(", countLeft + 1, res, max);
28             dfs(str.substring(1), subRes, countLeft, res, max);
29         } else if (str.charAt(0) == ')') {
30             if (countLeft > 0) {
31                 dfs(str.substring(1), subRes + ")", countLeft - 1, res, max);
32             }
33             dfs(str.substring(1), subRes, countLeft, res, max);
34         } else {
35             dfs(str.substring(1), subRes + str.charAt(0), countLeft, res, max);
36         }
37     }
38 } 

 

posted @ 2016-10-23 10:06  北叶青藤  阅读(265)  评论(0)  编辑  收藏  举报