題目描述：
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.
原始代碼略顯麻煩反肋，主要思路是不停改變最大與最小值:
int maxProfit(vector<int>& prices) {
if(prices.size()==0)
return 0;
int max = 0;
int buyin = prices[0];
int buyout = prices[0];
for(int i = 1;i<prices.size();i++){
if(prices[i]<buyin)
buyin = prices[i];
else{
buyout = prices[i];
if(buyout - buyin>max)
max = buyout -buyin;
}
}
return max;
}
改進(jìn)代碼簡潔明了嫉鲸，但思路相同：
int maxProfit(vector<int> &prices) {
int maxPro = 0;
int minPrice = INT_MAX;
for(int i = 0; i < prices.size(); i++){
minPrice = min(minPrice, prices[i]);
maxPro = max(maxPro, prices[i] - minPrice);
}
return maxPro;
}
