leetcode [#155]

目录

• 1. 题目
• 2. 解决方案
• 3. 注意事项

题目

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.

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解决方案

public class MinStack {
    private List<Integer> list;
    private int min = 2147483647;

    /** initialize your data structure here. */
    public MinStack() {
        list = new ArrayList<>();
    }

    public void push(int x) {
        list.add(x);
        if(list.size() == 1) {
            min = x;
        } else {
            if(x < min) min = x;
        }
    }

    public void pop() {
        list.remove(list.size() - 1);
        if(list.size() > 0){
            int tmp = list.get(0);
            for(int i = 1; i < list.size(); i++){
                if(list.get(i) < tmp){
                    tmp = list.get(i);
                }
            }
            min = tmp;
        }
    }

    public int top() {
        return list.get(list.size() - 1);
    }

    public int getMin() {
        return list.size() == 0 ? 0 : 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(); */
 * 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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注意事项

1. 按照堆栈的定义依次实现各方法。
2. push()操作时，压入元素的同时要更新min的值；getMin()操作时，直接返回min的值；pop()操作时，弹出元素的同时要记得更新min的值。