动态规划-121. Best Time to Buy and Sell Stock

题目

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: 5max. 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.

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题意解读：给定一个数组，数组的第i个元素是第i天的股票价格。如果你只允许进行一次买卖操作（买和买当做一次交易），设计一个算法寻找最大额收益

//d[i]代表[0,i]内交易股票的最大收益 minPrice表示［0，i）之间价格的最小值//状态转移方程 dp[i] = max(dp[i-1],prices[i]-minPrice)class Solution {    public int maxProfit(int[] prices) {        if(prices == null || prices.length==0) return 0;        int[] dp = new int[prices.length];        int minPrice = prices[0];        for(int i = 1; i < prices.length; i++){            minPrice = Math.min(minPrice,prices[i]);            dp[i] = Math.max(dp[i-1],prices[i]-minPrice);        }        return dp[prices.length-1];    }}