Time=O(1), space=(n)
Stack: 5 5 6 6 7 7 2 2
minStack: 5 5 5 5 5 5 2 2
class MinStack {
Deque<Integer> stack;
Deque<Integer> minStack;
/** initialize your data structure here. */
public MinStack() {
stack = new ArrayDeque<>();
minStack = new ArrayDeque<>();
}
public void push(int x) {
stack.push(x);
if (!minStack.isEmpty() && minStack.peek() < x) {
minStack.push(minStack.peek());
} else {
minStack.push(x);
}
}
public void pop() {
if (!stack.isEmpty()) {
stack.poll();
minStack.poll();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
Optimized usage of stack 2 (assuming there are lots of duplicates).
Time=O(1), space=(n)
Stack: 5 5 6 6 7 7 2 2
minStack: 5 5 2 2
class MinStack {
Deque<Integer> stack;
Deque<Integer> minStack;
/** initialize your data structure here. */
public MinStack() {
stack = new ArrayDeque<>();
minStack = new ArrayDeque<>();
}
public void push(int x) {
stack.push(x);
// when value <= current min value in stack, need to push the value to minStack
if (minStack.isEmpty() || minStack.peek() >= x) {
minStack.push(x);
}
}
public void pop() {
int x = stack.pop();
// When the popped value is the same as top value of minStack,
// the value need to be popped from minStack
if (x == minStack.peek()) {
minStack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
/**
* 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();
*/
TODO