题目: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.
题目解答:给定一组数表示每天的股价,求出只交易一次时可能获得的最大利益。其实就是获得当前位置的数字与之前出现过的最小股价的差最大的时候。
代码:
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() <= 0)
return 0;
int min = prices[0];
int profit = 0;
for(vector<int>::iterator vit = prices.begin();vit != prices.end();vit++)
{
profit = Max (*vit - min,profit);
if(*vit < min)
{
min = *vit;
}
}
return profit;
}
int Max(int a,int b)
{
return a > b ? a : b;
}
};