leetcode | 301. Remove Invalid Parentheses (BFS)

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题目

Input: "()())()"
Output: ["()()()", "(())()"]

Input: "(a)())()"
Output: ["(a)()()", "(a())()"]

Input: ")("
Output: [""]

题目描述：去除最小数量的无效括号，保证输入的括号有效，输出所有可能。

分析

此题适合BFS，按照删除的字符数量来分层，每个节点的子节点数量<=当前字符串的长度；因为涉及到最小数量，BFS可以知道满足条件的level，及时停止搜索；每层表示 - 去除${level}个括号的字符串；root是原字符串，第一层是去除一个括号的字符串；

解法

public class P301_RemoveInvalidParentheses {
    public List<String> removeInvalidParentheses(String s) {
        List<String> res = new LinkedList<>();
        Queue<String> queue = new LinkedList<>();
        Set<String> visited = new HashSet<>();
        queue.add(s);
        boolean found = false;
        while (!queue.isEmpty()) {
            String cur = queue.poll();
            if (isValidParentheses(cur)) {
                res.add(cur);
                found = true;
            }

            if (found) {
                continue;
            }

            for (int i = 0; i < cur.length(); i++) {
                if (cur.charAt(i) != '(' && cur.charAt(i) != ')') {
                    continue;
                }
                String sub = cur.substring(0, i) + (i == cur.length() - 1 ? "" : cur.substring(i + 1));
                if (visited.add(sub)) {
                    queue.offer(sub);
                }
            }
        }
        return res;
    }

    private boolean isValidParentheses(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            char x = s.charAt(i);
            if (x == '(') {
                count++;
            } else if (x == ')') {
                count--;
            }
            if (count < 0) {
                return false;
            }
        }
        return  count == 0;
    }
}