LeetCode[22] 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:

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

BackTracking

复杂度
O(N!),O(N)

思路
注意trick的地方,要限制左括号和右括号。每出现一次左括号,就相对于限定了,最多只能出现那么多右括号。所以,为了完成这种限定,用right - 1来控制。不然会有“(()))(”的情况出现。

代码

public List<String> generateParenthesis(int n) {
    List<String> res = new LinkedList<>();
    helper(res, 0, n, n, new StringBuilder());
    return res;
}

public void helper(List<String> res, int left, int right, int n, StringBuilder temp) {
    // Base case;
    if(left == n && right == n) {
        res.add(temp.roString());
    }
    if(left < n) {
        // 限制在当前这么多左括号的情况下,最多只能出现那么多的右括号。
        helper(res, left + 1, right - 1, n, temp.append("("));
        temp.deleteCharAt(temp.length() - 1);
    }
    if(right < n) {
        helper(res, left, right + 1, n, temp.append(")"));
        temp.deleteCharAt(temp.length() - 1);
    }
}

hellolittleJ17
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