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.
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分析：
思路首先：动态规划
遍历数组时记录当前价格以前的最小价格curMin，记录昨天能够获得的最大利润maxProfit
对于今天，为了能获得此刻的最大利润，显然只能卖，或者不做任何操作
如果不做任何操作，显然还是昨天maxProfit
如果卖掉今天天的股票，显然prices[i]-curMin
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<=1)
return 0;
int curMin=prices[0];
int maxProfit=0;
for(int i=1;i<prices.size();i++)
{
maxProfit=max(maxProfit,prices[i]-curMin);
curMin=min(curMin,prices[i]);//获得历史最小价格的股票
}
return maxProfit;
}
};
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原文地址：http://blog.csdn.net/ebowtang/article/details/50524380
原作者博客：http://blog.csdn.net/ebowtang
<LeetCode OJ> 121. Best Time to Buy and Sell Stock
原文：http://blog.csdn.net/ebowtang/article/details/50524380