LeetCode 131 Generate Parentheses
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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:
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
分析:
看见生成所有,基本可以确定是回溯问题,即DFS。
这个解法的特别之处在于,每次DFS时都会生成一个新的item实例,所以不需要递归结束时恢复现场。
public class Solution { public List<String> generateParenthesis(int n) { List<String> res = new ArrayList<String>(); dfs(res, new String(), n, n); return res; } public void dfs(List<String> res, String item, int leftNum, int rightNum){ //如果剩余的左括号数大于右括号数,是非法的 if(leftNum > rightNum) return; //递归终止 if(leftNum==0 && rightNum==0){ res.add(item); return; } if(leftNum > 0) //因为item+"("每次生成一个新的实例,所以不需要恢复现场 dfs(res, item+"(", leftNum-1, rightNum); if(rightNum > 0) dfs(res, item+")", leftNum, rightNum-1); }}
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