[LeetCode]20. Valid Parentheses

20. 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.

题意：
括号字符串的匹配。

栈的特性：
后进先出。

栈的头文件：

struct stack
{
char elem;

struct stack *next;

};
栈的大小：
在64位系统上：sizeof(char) = 1，sizeof(struct stack *) = 8，sizeof(struct stack) = 16;
在32位系统上：sizeof(char) = 1，sizeof(struct stack *) = 4，sizeof(struct stack) = 8;
64位系统的地址是64位的，即8字节；32位系统的地址是32的，即4字节。所以sizeof(struct stack *)分别是8字节和4字节。虽然sizeof(char)都为一字节。由于结构体存在字节对齐，所以sizeof取出的结构体大小是结构体最大元素的整数倍。所以结构体大小是16字节和8字节。

push入栈，pop出栈，top获取栈顶元素。getLen函数如果栈为空，则返回一；否则返回零。

思路：
如果元素是'('，'{'或'['则直接入栈；如果是元素是')'，则判断栈顶，如果栈顶元素是'('则'('出栈。'}'和']'一样处理。

struct stack
{
char elem;
struct stack *next;
};

struct stack
*push(struct stack *head, char c)
{
struct stack *node = NULL;
node = (struct stack *)malloc(sizeof(struct stack));
if ( node == NULL )
{
return NULL;
}

node->elem = c;
if ( head == NULL )
{
node->next = NULL;
head = node;
}
else
{
node->next = head;
}

return node;
}

char
top(struct stack *head)
{
if ( !head )
{
return 0;
}
return head->elem;
}

struct stack
*pop(struct stack *head)
{
if ( !head )
{
return NULL;
}
char val = head->elem;
struct stack *delete = head;
head = head->next;
free(delete);
return head;
}

int
getLen(struct stack *head)
{
return head == NULL ? 1 : 0;
}

bool isValid(char* s)
{
struct stack *head = NULL;
while ( *s )
{
if ( *s == '(' || *s == '{' || *s == '[' )
{
head = push(head, *s);
}

if ( *s == ')' )
{
if ( top(head) == '(' )
{
head = pop(head);
}
else
{
head = push(head, *s);
}
}

if ( *s == '}' )
{
if ( top(head) == '{' )
{
head = pop(head);
}
else
{
head = push(head, *s);
}
}

if ( *s == ']' )
{
if ( top(head) == '[' )
{
head = pop(head);
}
else
{
head = push(head, *s);
}
}
s++;
}

return getLen(head);
}