LeetCode:Valid Parentheses算法详解

算法题目：Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
The brackets must close in the correct order, “()” 、 “()[]{}” and “([])[{()}]”are all valid but “(]” and “([)]” are not.

大致意思：给定一个只包含’(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’这六个字符的字符串，要求判断字符串是否满足要求，要求是：以正确的次序关闭，例如：”()” 、 “()[]{}” 和 “([])[{()}]”是正确的，而”(]” and “([)]”是错误的。

思路：典型的递归算法题，将字符串分成两个串s1、s2，初始化s1=”“,s2=s，判断字符串s1的末尾字符与s2的开头字符是否配对，若配对，则将s1的末尾字符删除，s2的开头字符串删除，若不配对，则将s2[0]移到s1的末尾，然后继续递归下去。

bool MyValidCore(string s1,string s2){   if(s1.size()==0&&s2.size()==0)return true;   else if(s1.size()>0&&s2.size()==0)return false;   if(s1.size()==0)   {        s1+=s2[0];        s2=s2.substr(1);        return MyValidCore(s1,s2);   }   int l1=s1.size();   if((s1[l1-1]=='('&&s2[0]==')')||(s1[l1-1]=='['&&s2[0]==']')||(s1[l1-1]=='{'&&s2[0]=='}'))   {        s1=s1.substr(0,l1-1);        s2=s2.substr(1);   }   else   {        s1+=s2[0];        s2=s2.substr(1);   }   return MyValidCore(s1,s2);} bool isValid(string s) {   if(s.size()==0||s.size()%2==1)return false;   return MyValidCore("",s); }