valid parentheses

使用栈来解决：

/*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.*/#include <stdio.h>#include <stdlib.h>#include <assert.h>#define STACK_TYPE chartypedef struct STACK_NODE {STACK_TYPE value;struct STACK_NODE *next;} StackNode;STACK_TYPE top(StackNode *stack_top);int is_empty(StackNode *stack);void push2(StackNode **stack_top,STACK_TYPE value);void pop2(StackNode **stack_top);int isValid(char* s) {    StackNode *stack=NULL;char c;while(*s!='\0'){printf("%c\n",*s);switch(*s){case '(':case '[':case '{': push2(&stack,*s);break;case ')': if(is_empty(stack) || top(stack)!='('){//printf("top:%c\n",top(stack));return 0;}          else pop2(&stack);break;case ']':  if(is_empty(stack) || top(stack)!='['){//printf("top:%c\n",top(stack));return 0;}          else pop2(&stack);break;case '}':  if(is_empty(stack) || top(stack)!='{'){//printf("top:%c\n",top(stack));return 0;}          else pop2(&stack);break;default:;}//printf("top:%c\n",top(stack));s++;}return is_empty(stack);}int main(int argc,char **argv){ char *s="([])[()"; printf("%d\n",isValid(s));return 0;}void push2(StackNode **stack_top,STACK_TYPE value){StackNode *new_node=(StackNode*)malloc(sizeof(StackNode));assert(new_node!=NULL);new_node->value=value;new_node->next=*stack_top;*stack_top=new_node;}void pop2(StackNode **stack_top){StackNode *first_node;//assert(!is_empty(*stack_top));first_node=*stack_top;*stack_top=first_node->next;free(first_node);}STACK_TYPE top(StackNode *stack_top){return (stack_top)->value;}int is_empty(StackNode *stack){return (stack)==NULL;}

python:

# -*- coding: utf-8 -*-'''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.'''class Solution(object):    def isValid(self, s):        """        :type s: str        :rtype: bool        """        characters={')':'(',']':'[','}':'{'}        stack=[]        for c in s:        if c in characters.values():        stack.append(c)        elif c in characters.keys():        if stack==[] or characters[c]!=stack.pop():        return False        else:        return False        return stack==[]mystack=Solution()s='[](){][}'print mystack.isValid(s)