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.
一個循環，先存之前最小的值，再比較最大收益。循環完就找到了最大收益：
class Solution {
public:
int maxProfit(vector<int>& prices) {
int maxpro = 0;
int minpro = INT_MAX;
for(int i = 0; i < prices.size(); ++i){
minpro = min(minpro, prices[i]);
maxpro = max(maxpro, prices[i] - minpro);
}
return maxpro;
}
};