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 )
.
Example 1:
Input: "()())()" Output: ["()()()", "(())()"]
Example 2:
Input: "(a)())()" Output: ["(a)()()", "(a())()"]
Example 3:
Input: ")(" Output: [""]
思路:可以利用DFS或者BFS来解这道题,感觉还是BFS简单点,即对于从给定的字符串通过移除 ( 或 ) 来构造所有可能的合法串,如果合法就加入到set集合中,不合法到到下一轮的BFS中。
public class Solution { public List<String> removeInvalidParentheses(String s) { List<String> res = new ArrayList<>(); // sanity check if (s == null) return res; Set<String> visited = new HashSet<>(); Queue<String> queue = new LinkedList<>(); // initialize queue.add(s); visited.add(s); boolean found = false; while (!queue.isEmpty()) { s = queue.poll(); // 如果当前层次中有合法解的话,只需要将当前层次中的字符串全部弹出判断是否合法,停止BFS,这样保证所得到的合法字符串是移除最少字符得到的 if (isValid(s)) { // found an answer, add to the result res.add(s); found = true; } if (found) continue; // generate all possible states for (int i = 0; i < s.length(); i++) { // we only try to remove left or right paren if (s.charAt(i) != ‘(‘ && s.charAt(i) != ‘)‘) continue; String t = s.substring(0, i) + s.substring(i + 1); if (!visited.contains(t)) { // for each state, if it‘s not visited, add it to the queue queue.add(t); visited.add(t); } } } return res; } // helper function checks if string s contains valid parantheses boolean isValid(String s) { int count = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == ‘(‘) count++; if (c == ‘)‘ && count-- == 0) return false; } return count == 0; } }
原文地址:https://www.cnblogs.com/f91og/p/9668209.html
时间: 2024-07-29 21:44:04