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: 0In this case, no transaction is done, i.e. max profit = 0.
解题思路,定义数组dp,dp[i]表示在 i 处抛售股票的最大获利,该最大获利表示的是从prices数组0到 i 处的最大获利,由该处的价格减去prices数组中从0到 i -1 处的最小值。得到每个地方的最大获利后,在循环便利dp数组,就可以得到相对整个数组来说的最大获利。
public class Solution { public int maxProfit(int[] prices) { if(prices.length<=1){ return 0; }else{ int min=prices[0]; int [] dp= new int[prices.length]; dp[0]=0; for(int i=1;i<prices.length;i++){ dp[i]=prices[i]-min; if(prices[i]<min){ min = prices[i]; } } int max=dp[0]; for(int i=1;i<dp.length;i++){ if(dp[i]>max){ max=dp[i]; } } max=0>max?0:max; return max; } } }