Description
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
Two stacks
用一个minStack来辅助记录，当push时与minStack栈顶进行比较，比栈顶值小才入栈，维持minStack中栈顶到栈底是一个递增的关系。pop的时候也需要与minStack栈顶比较。
class MinStack {
private Stack<Integer> stack;
private Stack<Integer> minStack;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack();
minStack = new Stack();
}
public void push(int x) {
stack.push(x);
if (x <= getMin()) {
minStack.push(x);
}
}
public void pop() {
if (stack.isEmpty()) return;
int val = stack.pop();
if (val == getMin()) {
minStack.pop();
}
}
public int top() {
if (stack.isEmpty()) {
return Integer.MAX_VALUE;
}
return stack.peek();
}
public int getMin() {
if (minStack.isEmpty()) return Integer.MAX_VALUE;
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();
*/
