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:
代码语言：javascript复制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:
代码语言：javascript复制Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.买股票，低买，高卖，问最大收益。
DP，记录当前位置和之前的最小值，每个位置的价格和之前最小值的差价表示当前位置卖的最大收益。
代码语言：javascript复制class Solution {
public:
int maxProfit(vector<int>& prices) {
int res=0;
for(int i=1;i<prices.size();i )
{
res=max(res,prices[i]-prices[i-1]);
prices[i]=min(prices[i],prices[i-1]);
}
return res;
}
};
