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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())()"]
")(" -> [""]
看到parenthese的问题，第一反应是用栈。这题要求minimum number，所以想到用BFS遍历解空间树。
思路为：
层次依次为删除0个元素，1个元素，2个元素。。。
层次遍历所有的可能。如果有一种可能是valid，那么不再遍历下面的层。
1 public class Solution { 2 public List<String> removeInvalidParentheses(String s) { 3 Set<String> visited = new HashSet<String>(); 4 List<String> result = new ArrayList<String>(); 5 List<String> current = new ArrayList<String>(); 6 List<String> next; 7 current.add(s); 8 boolean reached = false; 9 // BFS 10 while (!current.isEmpty()) { 11 next = new ArrayList<String>(); 12 for (String prev : current) { 13 visited.add(prev); 14 // If valid 15 if (isValid(prev)) { 16 reached = true; 17 result.add(prev); 18 } 19 20 // If not reached, then delete 21 if (!reached) { 22 for (int i = 0; i < prev.length(); i++) { 23 char tmp = prev.charAt(i); 24 if (tmp != ‘(‘ && tmp != ‘)‘) { 25 continue; 26 } 27 String newStr = prev.substring(0, i) + prev.substring(i + 1); 28 if (!visited.contains(newStr)) { 29 next.add(newStr); 30 visited.add(newStr); 31 } 32 } 33 } 34 } 35 if (reached) { 36 break; 37 } 38 current = next; 39 } 40 return result; 41 } 42 43 private boolean isValid(String s) { 44 Deque<Integer> stack = new LinkedList<Integer>(); 45 for (int i = 0; i < s.length(); i++) { 46 char cur = s.charAt(i); 47 if (cur != ‘(‘ && cur != ‘)‘) { 48 continue; 49 } 50 if (cur == ‘(‘) { 51 stack.push(i); 52 } 53 if (cur == ‘)‘) { 54 if (stack.isEmpty()) { 55 return false; 56 } else { 57 stack.pop(); 58 } 59 } 60 } 61 return stack.isEmpty(); 62 } 63 }
事实上，我们可以不用栈，用一个count来检测是否是valid parenthese.
1 private boolean isValid(String s) { 2 int count = 0; 3 for (int i = 0; i < s.length(); i++) { 4 char cur = s.charAt(i); 5 if (cur != ‘(‘ && cur != ‘)‘) { 6 continue; 7 } 8 if (cur == ‘(‘) { 9 count++; 10 } 11 if (cur == ‘)‘) { 12 count--; 13 if (count < 0) { 14 return false; 15 } 16 } 17 } 18 return count == 0; 19 }
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4944260.html
