14.Best Time to Buy and Sell Stock

标签：ted   ase   bsp   toc   most   imu   output   array   maximum   

题目描述：

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.


同样是动态规划的题目，但这次是求元素间最大的差，且必须是后面的元素减去前面元素所得的最大差（具有潜在的时间属性），这里其实一个用遍历就能解决

class Solution {
public:
　　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;
　　}
};

14.Best Time to Buy and Sell Stock

标签：ted   ase   bsp   toc   most   imu   output   array   maximum   

原文地址：http://www.cnblogs.com/sarahp/p/7080929.html