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包含min函数的栈

[编程题]包含min函数的栈
  • 热度指数:657860 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 256M,其他语言512M
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定义栈的数据结构,请在该类型中实现一个能够得到栈中所含最小元素的 min 函数,输入操作时保证 poptopmin 函数操作时,栈中一定有元素。

此栈包含的方法有:
push(value):将value压入栈中
pop():弹出栈顶元素
top():获取栈顶元素
min():获取栈中最小元素

数据范围:操作数量满足 ,输入的元素满足
进阶:栈的各个操作的时间复杂度是 ,空间复杂度是

示例:
输入:    ["PSH-1","PSH2","MIN","TOP","POP","PSH1","TOP","MIN"]
输出:    -1,2,1,-1
解析:
"PSH-1"表示将-1压入栈中,栈中元素为-1
"PSH2"表示将2压入栈中,栈中元素为2,-1
MIN”表示获取此时栈中最小元素==>返回-1
"TOP"表示获取栈顶元素==>返回2
"POP"表示弹出栈顶元素,弹出2,栈中元素为-1
"PSH1"表示将1压入栈中,栈中元素为1,-1
"TOP"表示获取栈顶元素==>返回1
MIN”表示获取此时栈中最小元素==>返回-1

示例1

输入

 ["PSH-1","PSH2","MIN","TOP","POP","PSH1","TOP","MIN"]

输出

-1,2,1,-1
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import java.util.Stack;
import java.util.Arrays;
public class Solution {
/*借用辅助栈存储min的大小,自定义了栈结构
*/
    private int size;
    private int min = Integer.MAX_VALUE;
    private Stack<Integer> minStack = new Stack<Integer>();
    private Integer[] elements = new Integer[10];
    public void push(int node) {
        ensureCapacity(size+1);
        elements[size++] = node;
        if(node <= min){
            minStack.push(node);
            min = minStack.peek();
        }else{
            minStack.push(min);
        }
    //    System.out.println(min+"");
    }

    private void ensureCapacity(int size) {
        // TODO Auto-generated method stub
        int len = elements.length;
        if(size > len){
            int newLen = (len*3)/2+1; //每次扩容方式
            elements = Arrays.copyOf(elements, newLen);
        }
    }
    public void pop() {
        Integer top = top();
        if(top != null){
            elements[size-1] = (Integer) null;
        }
        size--;
        minStack.pop();    
        min = minStack.peek();
    //    System.out.println(min+"");
    }

    public int top() {
        if(!empty()){
            if(size-1>=0)
                return elements[size-1];
        }
        return (Integer) null;
    }
    public boolean empty(){
        return size == 0;
    }

    public int min() {
        return min;
    }
} 

编辑于 2015-06-19 17:57:04 回复(57)