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 elementval
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 1:
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:
-231 <= val <= 231 - 1
- Methods
pop
,top
andgetMin
operations will always be called on non-empty stacks. - At most
3 * 104
calls will be made topush
,pop
,top
, andgetMin
.
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class MinStack { | |
public: | |
// keep with min value | |
stack<pair<int, int>> st; | |
MinStack() { | |
} | |
void push(int val) { | |
if(st.empty()) st.push({val, val}); | |
else st.push({val, min(val, st.top().second)}); | |
} | |
void pop() { | |
st.pop(); | |
} | |
int top() { | |
return st.top().first; | |
} | |
int getMin() { | |
return st.top().second; | |
} | |
}; | |
/** | |
* Your MinStack object will be instantiated and called as such: | |
* MinStack* obj = new MinStack(); | |
* obj->push(val); | |
* obj->pop(); | |
* int param_3 = obj->top(); | |
* int param_4 = obj->getMin(); | |
*/ |
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