15算法课程 121. Best Time to Buy and Sell Stock

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: 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.

solution:

找到交易价格中最大的股票利润，假设最低点为prices[i]，最高点为prices[j]，最大利润maxSum=prices[j]-prices[i]，假设i和j之间有k个中间价格，则maxSum=prices[j]-prices[x1]+prices[x1]-prices[x2]+...+prices[xk]-prices[i]，可以用动态规划解决，设向量sum记录相邻两个价格之间的最大利润，则sum[i]=max(prices[i]-prices[i-1]，sum[i-1]+prices[i]-prices[i-1])，即是按照i-1的价格出售获利大，还是按照sum[i-1]时的价格出售获利大，同时更改maxSum记录sum向量中最大的元素。

code:

class Solution {public:    int maxProfit(vector<int>& prices) {        vector<int> sum(prices.size(),0);        int maxSum=0;        for(int i=1;i<prices.size();++i){            int temp=prices[i]-prices[i-1];            sum[i]=max(temp,sum[i-1]+temp);            maxSum=max(maxSum,sum[i]);        }        return maxSum;    }};