LeetCode[155] 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.

分析:

Java中本来就有Stack。本题跟普通Stack的区别是需要记住最小值,关键在于push和pop方法。在push时,若x比当前的min更小,则先push值min,再push值x,并令min = x。在pop时,若该值为min值,需要pop两次,并且让第二次pop的值为min值。push和pop相呼应。

代码:

public class MinStack {
Stack<Integer> stack = new Stack<Integer>();
int min = Integer.MAX_VALUE;
/** initialize your data structure here. */
public MinStack() {
}
public void push(int x) {
if (x <= min) {
stack.push(min);
min = x;
}
stack.push(x);
}
public void pop() {
if (stack.pop() == min) {
min = stack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return 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();
*/

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