LeetCode-Remove Invalid Parentheses

Posted 2020-11-21 IncredibleThings

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: [""]

class Solution {
    public List<String> removeInvalidParentheses(String s) {
        Set<String> visited = new HashSet<>();
        List<String> res = new ArrayList<>();
        List<String> current = new ArrayList<>();
        List<String> next;
        boolean reached = false;
        
        current.add(s);
        
        while(!current.isEmpty()){
            next = new ArrayList<>();
            for(String prv : current){
                visited.add(prv);
                //if valid
                if(isValid(prv)){
                    reached = true;
                    res.add(prv);
                }
                 //not valid
                if(!reached){
                    for(int i=0; i<prv.length(); i++){
                        char tmp = prv.charAt(i);
                        if(tmp == ‘(‘ || tmp == ‘)‘){
                            String newString = prv.substring(0,i)+ prv.substring(i+1);
                            if(!visited.contains(newString)){
                                next.add(newString);
                                visited.add(newString);
                            }
                        }
                    }
                }
            }
            if(reached){
                break;
            }
            current = next;
        }
        return res;
        
    }
    
    
    private boolean isValid(String s){
        Stack<Character> stack = new Stack<>();
        for(int i=0; i<s.length(); i++){
            char c = s.charAt(i);
            if(c == ‘(‘){
                stack.push(c);
            }
            else if (c == ‘)‘){
                if(stack.isEmpty()){
                    return false;
                }
                else{
                    stack.pop();
                }
            }
        }
        return stack.isEmpty();
    }
}

　　

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。