LeetCode刷题之Min Stack

Problem

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.
My Solution
class MinStack {

    public int[] data;
    public int min = Integer.MAX_VALUE;
    public int top = -1;

    /** initialize your data structure here. */
    public MinStack() {
        data = new int[10000];
    }

    public void push(int x) {
        if (x < min) {
           min = x;
        }
        ++top;
        if (top < data.length - 1) {
            data[top] = x;
        }
    }

    public void pop() {
        if (top >= 0) {
            top--;
        }
        int t = top;
        min = Integer.MAX_VALUE;
        while (t != -1) {
            if (data[t] < min) {
               min = data[t];
            }
            t--;
        }
    }

    public int top() {
        if (top >= 0 && top <= data.length - 1) {
           return data[top];
        } else {
            return 0;
        }
    }

    public int getMin() {
       return min;
    }
}
Great Solution
public class MinStack {
    long min;
    Stack<Long> stack;

    public MinStack(){
        stack=new Stack<>();
    }
    
    public void push(int x) {
        if (stack.isEmpty()){
            stack.push(0L);
            min=x;
        }else{
            stack.push(x-min);//Could be negative if min value needs to change
            if (x<min) min=x;
        }
    }

    public void pop() {
        if (stack.isEmpty()) return;
        
        long pop=stack.pop();
        
        if (pop<0)  min=min-pop;//If negative, increase the min value
        
    }

    public int top() {
        long top=stack.peek();
        if (top>0){
            return (int)(top+min);
        }else{
           return (int)(min);
        }
    }

    public int getMin() {
        return (int)min;
    }
}
    原文作者:Gandalf0
    原文地址: https://www.jianshu.com/p/61005a18dae9
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