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22. generate parentheses

admin 12月 10, 2020 0

问题

https://leetcode.com/problems/generate-parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

思路

方法一：遍历所有可能的组合情况，对于符合条件的加入到结果中，不满足的忽略；
方法二：利用DFS的思想，递归左括号小于右括号且用完了所有的括号的所有情况，直接加入到结果中；

代码

public boolean (String s) {
    Stack<Character> stack = new Stack<Character>();
    for (char c : s.toCharArray()) {
        if (c == '(') {
            stack.push(')');
        } else if (c == '{') {
            stack.push('}');
        } else if (c == '[') {
            stack.push(']');
        } else if (stack.isEmpty() || stack.pop() != c) {
            return false;
        }
    }
    return stack.isEmpty();
}

public List<String> generateParenthesis(int n) {
    List<String> res = new ArrayList<>();
    if (n <= 0) {
        return res;
    }
    genParenthesis(res, new StringBuilder(), 0, n * 2);
    return res;
}

private void genParenthesis(List<String> res, StringBuilder sb, int index, int n) {
    if (sb.length() == n) {
        if (isValid(sb.toString())) {
            res.add(sb.toString());
        }
        return;
    }
    sb.append('(');
    genParenthesis(res, sb, index + 1, n);
    sb.deleteCharAt(sb.length() - 1);
    sb.append(')');
    genParenthesis(res, sb, index + 1, n);
    sb.deleteCharAt(sb.length() - 1);
}


public void dfs(List<String> res, String cur, int l, int r) {
    System.out.println("res:" + cur);
    if(l == 0 && r == 0) {
        res.add(cur);
        return;
    }
    if(l > 0) {
        dfs(res, cur + "(", l - 1, r);
    }
    if(l < r) {
        dfs(res, cur + ")", l, r -1);
    }

}

public List<String> generateParenthesis2(int n) {
    List<String> res = new ArrayList<>();
    if(n <= 0) {
        return res;
    }
    dfs(res, "", n , n);
    return res;
}