121. Best Time to Buy and Sell Stock

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.

解答：

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

动态规划法:
用sell[i]表示到第i天为止选择是否卖股票的最大收益，第一天为0；
用buy[i]表示到第i天时购买股票的最低价格，第一天为-prices[0]；
sell[i]的值，要么跟昨天一样，要么今天把这支股票卖掉所得的收益，取二者最大值；
sell[i] = max(sell[i-1], buy[i-1]+prices[i]);
buy[i]的值，即要么跟昨天时候一样（没有买今天的股票），要么买的是今天的股票，取二者最大值，为负值。
buy[i] = max(buy[i-1], -prices[i]);
最后的收益即为sell[prices.size()-1].