---
title: 'LeetCode 155. Min Stack'
disqus: hackmd
---
# LeetCode 155. Min Stack
## Description
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack class:
* MinStack() initializes the stack object.
* void push(int val) pushes the element val onto the stack.
* void pop() removes the element on the top of the stack.
* int top() gets the top element of the stack.
* int getMin() retrieves the minimum element in the stack.
## Example
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
## Constraints
-2^31^ <= val <= 2^31^ - 1
Methods pop, top and getMin operations will always be called on non-empty stacks.
At most 3 * 10^4^ calls will be made to push, pop, top, and getMin.
## Answer
此題實作stack,先開一個固定空間,push就塞進去,top++。而pop就 top--即可。
```Cin=
//2021_11_22
typedef struct {
int *nums;
int *min;
int size;
int top;
} MinStack;
MinStack* minStackCreate() {
MinStack *obj = (MinStack*)malloc(sizeof(MinStack));
obj->size = 1000;
obj->nums = (int*)malloc(sizeof(int)*obj->size);
obj->min = (int*)malloc(sizeof(int)*obj->size);
obj->top = 0;
return obj;
}
void minStackPush(MinStack* obj, int val) {
if(obj->top < obj->size){
obj->nums[obj->top] = val;
if(obj->top == 0){obj->min[obj->top] = val;}
else{obj->min[obj->top] = obj->min[obj->top - 1] < val ? obj->min[obj->top - 1] : val;}
(obj->top)++;
}
}
void minStackPop(MinStack* obj) {
if(obj->top > 0){
(obj->top)--;
}
}
int minStackTop(MinStack* obj) {
return obj->nums[obj->top - 1];
}
int minStackGetMin(MinStack* obj) {
return obj->min[obj->top - 1];
}
void minStackFree(MinStack* obj) {
free(obj->nums);
free(obj->min);
free(obj);
}
```
## Link
https://leetcode.com/problems/min-stack/
###### tags: `Leetcode`