LeetCode 121: Best Time to Buy and Sell Stock

Posted 2020-09-10 嗡嘛呢巴美吽舍

题目如下：

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

思路：

遍历数组，找当前最小值curMin，结果max一定是当前最小值之后一个数和它的差。

遍历过程中不断更新curMin和max。

复杂度O(N)；

 

本题代码：

import java.util.Scanner;

/**
 * Created by yuanxu on 17/4/12.
 */
public class DP121 {

    public static int maxProfit(int[] prices) {
        if (prices.length == 0) {
            return 0;
        }
        int max = 0;
        int curMin = prices[0];
        for (int i=1; i<prices.length; i++) {
            if (prices[i] > curMin) {
                max = (prices[i] - curMin) > max ? (prices[i] - curMin) : max;
            } else {
                curMin = prices[i];
            }
        }
        return max;
    }


    public  static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        String inputString = scanner.nextLine();
        String stringArray[] = inputString.substring(1,inputString.length()-1).split(", ");
        int prices[] = new int[stringArray.length];
        for(int i=0; i<stringArray.length; i++) {
            prices[i] = Integer.parseInt(stringArray[i]);
        }
        System.out.println(maxProfit(prices));
    }



}