这里，我提供一种用栈来解决的方法：

思路：栈的结构是先进后出，这样我们就可以模拟栈结构了，如果是‘（’、‘{’、‘[’任何一种，直接push进栈就可以了，如果是‘}’、‘）’、‘]’任何一种就开始判断，看栈pop的是否和对应的字符匹配。

 

 

 下面是源码：

typedef char STDateType;
typedef struct Stack
{STDateType* a;int top;int capacity;
}Stack;void StackInit(Stack* ps);
void StackPush(Stack* ps, STDateType x);
void StackPop(Stack* ps);
STDateType StackTop(Stack* ps);
int StackSize(Stack* ps);
bool StackEmpty(Stack* ps);
void StackDestroy(Stack* ps);void StackInit(Stack* ps)
{assert(ps);ps->a = NULL;ps->capacity = ps->top = 0;
}void StackPush(Stack* ps, STDateType x)
{assert(ps);if (ps->top == ps->capacity){int newcapacity = ps->capacity == 0 ? 4 : ps->capacity * 2;STDateType* tmp = (STDateType*)realloc(ps->a, sizeof(STDateType) * newcapacity);if (tmp == NULL){perror("realloc fail");exit(-1);}ps->a = tmp;ps->capacity = newcapacity;}ps->a[ps->top] = x;ps->top++;
}void StackPop(Stack* ps)
{assert(ps);--ps->top;
}STDateType StackTop(Stack* ps)
{assert(ps);return ps->a[ps->top - 1];
}int StackSize(Stack* ps)
{assert(ps);return ps->top;
}bool StackEmpty(Stack* ps)
{assert(ps);return ps->top == 0;
}void StackDestroy(Stack* ps)
{assert(ps);free(ps->a);ps->a = NULL;ps->top = ps->capacity = 0;
}bool isValid(char * s){Stack st;StackInit(&st);char top;while(*s){if((*s == '(') || (*s == '[') || (*s == '{')){StackPush(&st,*s);}else{if(StackEmpty(&st)){StackDestroy(&st);return false;}top = StackTop(&st);StackPop(&st);if((*s == ']' && top != '[')|| (*s == '}' && top != '{')|| (*s == ')' && top != '(')){StackPop(&st);StackDestroy(&st);return false;  	}}++s;}bool ret = StackEmpty(&st);StackDestroy(&st);return ret;
}