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.
vector下标，初始化
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class
Solution {public: int
maxProfit(vector<int> &prices) { if(prices.size() == 0)return
0; int
s = prices.size(); vector<int> min(s,prices[0]); vector<int> max(prices.size(),prices[prices.size()-1]); //min.push_back(prices[0]); for(int
i = 1 ; i < s ; i++) if(min[i - 1] > prices[i]) min[i] = prices[i]; else
min[i] = min[i-1]; //max.push_back(prices[s-1]); for(int
j = s-2 ; j >= 0 ; j --) if(prices[j] > max[j+1]) max[j] = prices[j]; else
max[j] = max[j+1]; int
pro = 0 ; for(int
i = 0 ; i < s ;i++)if(pro < max[i] - min[i]) pro = max[i] - min[i]; return
pro; }}; |
更简单的方法：
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class
Solution {public: int
maxProfit(vector<int> &prices) { // Start typing your C/C++ solution below // DO NOT write int main() function if
(prices.size() == 0) { return
0; } int
min = prices[0], profit = 0; for
(int i = 1; i < prices.size(); i++) { profit = prices[i] - min > profit ? prices[i] - min : profit; min = prices[i] < min ? prices[i] : min; } return
profit; }}; |
空间上更优
Best Time to Buy and Sell Stock,布布扣,bubuko.com
Best Time to Buy and Sell Stock
原文：http://www.cnblogs.com/pengyu2003/p/3580404.html