LeetCode:Remove Invalid Parentheses

Posted 2020-07-26 walker lee

Remove Invalid Parentheses

Total Accepted: 14168 Total Submissions: 43620 Difficulty: Hard

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())()"]
")(" -> [""]

Credits:
Special thanks to @hpplayer for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

Hide Tags

 Depth-first Search Breadth-first Search

Hide Similar Problems

 (E) Valid Parentheses

思路：

BFS

java code:

public class Solution {
    public List<String> removeInvalidParentheses(String s) {
        
        List<String> ans = new ArrayList<>();
        if(s == null) return ans;
        
        Set<String> visited = new HashSet<>();
        Queue<String> queue = new LinkedList<>();
        queue.offer(s);
        
        boolean found = false;
        while(!queue.isEmpty()) {
            
            String tmp = queue.poll();
            if(isValid(tmp)) {
                ans.add(tmp);
                found = true;
            }
            if(found) continue;
            
            for(int i = 0;i < tmp.length();i++) { //将各个（去掉i字符的）子串入队
                if(tmp.charAt(i) != '(' && tmp.charAt(i) != ')') continue;
                String sub = tmp.substring(0,i) + tmp.substring(i+1);
                
                if(!visited.contains(sub)) {
                    queue.offer(sub);
                    visited.add(sub);
                }
            }
        }
        return ans;
    }
    
    // 自定义函数：检查是否合法
    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 == ')') {
                if(count==0) return false;
                count--;
            }
        }
        return count == 0;
    }
}