最小栈问题

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。

• push(x) -- 将元素 x 推入栈中。
• pop() -- 删除栈顶的元素。
• top() -- 获取栈顶元素。
• getMin() -- 检索栈中的最小元素。

示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.getMin();   --> 返回 -2.

typedef struct node {
    int num;
    struct node *next;
}Node;

typedef struct stack {
    struct node *top;
    struct node *min;
} MinStack;

/** initialize your data structure here. */

MinStack* minStackCreate() {
    MinStack *new = malloc(sizeof(MinStack));
    new->top = NULL;
    new->min = NULL;
    return new;
}

void minStackPush(MinStack* obj, int x) {
    struct node *newnode = malloc(sizeof(struct node));
    newnode->num = x;
    
    newnode->next = obj->top;
    obj->top = newnode;
    
    if (!obj->min) {
        obj->min = newnode;
    }
    else {
        if (newnode->num < obj->min->num) {
            obj->min = newnode;
        }
    }        
}

void minStackPop(MinStack* obj) {
    struct node *popnode = obj->top;
    struct node *minnode;
    struct node *tmp;
    if (popnode) {
        if (obj->min == popnode) {
            tmp = popnode->next;
            minnode = tmp;
            while(tmp && tmp->next) {
                tmp = tmp->next;
                if (tmp->num < minnode->num)
                    minnode = tmp;
            }
            obj->min = minnode;
        }

        obj->top = popnode->next;
        free(popnode);
        popnode = NULL;
    }
    
    
}

int minStackTop(MinStack* obj) {
    return obj->top->num;
}

int minStackGetMin(MinStack* obj) {
    return obj->min->num;
}

void minStackFree(MinStack* obj) {
    struct node *tmp = obj->top;
    
    while(tmp != NULL) {
        obj->top = tmp->next;
        free(tmp);
        tmp = obj->top;
    }
    
    free(obj);
    obj = NULL;
}