【Leetcode-Medium-22】Generate Parentheses

题目

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

backtracking solution，回溯The idea here is to only add '(' and ')' that we know will guarantee us a solution (instead of adding 1 too many close).Once we add a '(' we will then discard it and try a ')' which can only close a valid '('. Each of these steps are recursively called.

程序

import java.util.*;class Solution {    /**    backtracking solution，回溯    The idea here is to only add '(' and ')' that we know will guarantee us a solution (instead of adding 1 too many close).    Once we add a '(' we will then discard it and try a ')' which can only close a valid '('.     Each of these steps are recursively called.    */    public List<String> generateParenthesis(int n) {        List<String> list = new ArrayList<>();        backtrack(list, "", 0, 0, n);        return list;    }    private void backtrack(List<String> list, String str, int open, int close, int max){        if (str.length() == 2*max){            list.add(str);            return;        }        if (open < max)            backtrack(list, str+"(", open+1, close, max);        if (close < open)            backtrack(list, str+")", open, close+1, max);    }    //------------------------------------------------    // Time limit exceeded    /*    思路：        排列组合出所有的字符串        验证每一个字符串是否合格    */    public List<String> generateParenthesis0(int n) {        char[] chs = new char[n*2];        for (int i = 0; i < n; i ++){            chs[i] = '(';            chs[n+i] = ')';        }        Set<String> set = new HashSet<>();        permutate(chs, 0, set);        return new ArrayList<String>(set);    }    // 字符串全排列    private void permutate(char[] chs, int start, Set<String> set){        if (start == chs.length){            boolean bool = validate(chs);            if (bool)                set.add(new String(chs));        }        for (int i = start; i < chs.length; i ++){            char temp = chs[start];            chs[start] = chs[i];            chs[i] = temp;            permutate(chs, start+1, set);            temp = chs[start];            chs[start] = chs[i];            chs[i] = temp;        }    }    private boolean validate(char[] chs){        LinkedList<Character> stack = new LinkedList<>();        for (char ch : chs){            if (ch == ')') {                if (stack.isEmpty() || stack.pop() != '(') return false;            } else {                stack.push(ch);            }        }        return stack.isEmpty();    }   }

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