121 Best Time to Buy and Sell Stock

121 Best Time to Buy and Sell Stock JAVA实现

• 题目描述

  Say you have an array for which the i*th 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.

  • 题目大意：给定一个数组，数组中的第i个元素，是第i天股票的价格，最多允许完成一次交易（买一次，卖一次），设计一个算法去寻找最大收益。

  • 思路：dp，遍历数组，寻找当前遍历过程中的最小值，如果当前的值比最小值大，则可获得收益，收益为（当前值-最小值）获取收益后与之前收益进行对比，获取收益较大的值，如果当前值比最小值小，更新最小值，遍历数组，寻找最大收益。

  • 代码实现：

  package Array;
  
  public class Solution {
      public int maxProfit(int[] prices) {
          if (prices == null || prices.length == 0) {
              return 0;
          }
  //        最大收益
          int maxProfit = 0;
  //        最小值
          int min = prices[0];
          for(int i=1;i<prices.length;i++) {
  //            当前值比之前的最小值大，则可假设在之前最小值时买入，在当前卖出，可以获得收益
              if (prices[i] > min) {
                  maxProfit = Math.max(maxProfit, prices[i] - min);
              } else {
  //                如果当前值小于等于之最小值，则更新最小值，作为买入
                  min = prices[i];
              }
          }
          return maxProfit;
      }
      public static void main(String[] args) {
  
      }
  
  }