Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
Solution
class MinStack:
def __init__(self):
self.q = []
# @param x, an integer
# @return an integer
def push(self, x):
curMin = self.getMin()
if curMin == None or x < curMin:
curMin = x
self.q.append((x, curMin));
# @return nothing
def pop(self):
self.q.pop()
# @return an integer
def top(self):
if len(self.q) == 0:
return None
else:
return self.q[len(self.q) - 1][0]
# @return an integer
def getMin(self):
if len(self.q) == 0:
return None
else:
return self.q[len(self.q) - 1][1]
public class MinStack {
private long min;
Stack<Long> stack;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack<>();
}
public void push(int x) {
if(stack.isEmpty()) {
stack.push(0L);
min = x;
} else {
stack.push(x - min);
if(x < min) min = x;
}
}
public void pop() {
if(stack.isEmpty()) return;
long pop = stack.pop();
if(pop < 0)
min = min - pop;
}
public int top() {
long top = stack.peek();
if (top>0){
return (int)(top+min);
}else{
return (int)(min);
}
}
public int getMin() {
return (int) min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/