[C语言][LeetCode][20]Valid Parentheses

题目

Valid Parentheses
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.

标签

Stack、String

难度

简单

分析

题目意思是给定一个字符串，仅包含’(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’，确定字符串是否有效。

实现思路是利用栈，遍历字符串，如果遇到’(‘, ‘{‘, ‘[‘，则入栈 ，如果遇到’)’, ‘}’, ‘]’，则比较栈顶元素，如果是对应的字符，则出栈，否则返回false

C代码实现

typedef char ElemType;typedef struct STACK_T {    ElemType value;    struct STACK_T * next;}STACK;typedef struct STACK_T NODE;STACK * stack = NULL;STACK * stack_init(void){    STACK * stack = (STACK *)malloc(sizeof(STACK));    if(!stack)    {        printf("malloc stack fail\n");        return NULL;    }    memset(stack, 0, sizeof(STACK));    stack->next = NULL;    return stack;}bool stack_is_empty(STACK * stack){    return (stack->next == NULL);}ElemType stack_pop(STACK * stack){    ElemType retValue;    STACK * temp = NULL;    if(!stack_is_empty(stack))    {        temp = stack->next;        stack->next = stack->next->next;        retValue = temp->value;        free(temp);    }    else    {        printf("stack is empty\n");        return 0;    }    return retValue;}int stack_push(STACK * stack, ElemType ele){    NODE * node = (NODE *)malloc(sizeof(NODE));    //memcpy(&node->Element, ele, sizeof(ElemType));    node->value = ele;    node->next = stack->next;    stack->next = node;    return 0;}ElemType stack_top(STACK * stack){    if(!stack_is_empty(stack))    {        return stack->next->value;    }    return (ElemType)(-1);}bool isValid(char* s) {    char * p = s;    if(!p)        return false;    if(*(p+1) == '\0')        return false;    stack = stack_init();    while(*p != '\0')    {        if( (*p == '(') ||(*p == '{') || (*p == '[') )            stack_push(stack, *p);        else if(*p == ')')        {            if('(' != stack_top(stack))                return false;            else                 stack_pop(stack);        }        else if(*p == '}')        {            if('{' != stack_top(stack))                return false;            else                 stack_pop(stack);        }        else if(*p == ']')        {            if('[' != stack_top(stack))                return false;            else                 stack_pop(stack);        }        else             return false;        p = p + 1;    }    if(true == stack_is_empty(stack))        return true;    else        return false;}