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:
"((()))", "(()())", "(())()", "()(())", "()()()"
public class Solution {
public List<String> generateParenthesis(int n) {
int left=0;
int right=0;
String str="";
List<String> list =new ArrayList<String>();
gen(left, right, n, str, list);
return list;
}
public void gen(int left, int right, int n, String str, List<String> list){
if( left==n && right==n){
list.add(str);
return;
}
if( left<n && left==right){
gen(left+1, right, n,str+"(", list);
}
else if( left<n && left>right ){
gen(left+1, right, n,str+"(", list);
gen(left, right+1, n,str+")", list);
}
else if(left==n && right<n ){
gen(left, right+1, n,str+")", list);
}
}
}主要用到了递归的思想,递归时保证字符串中左括号数大于等于右括号数即可。Generate Parentheses,布布扣,bubuko.com
原文地址:http://blog.csdn.net/dutsoft/article/details/37700647