#week8
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.
分析：
动态规划
每个地方的最大profit为本身减去前面最小的（当前面最小的>=本身时为0）
状态转换方程：
low[i] = (low[i-1] < prices[i])? low[i-1]:prices[i];
f[i] = (low[i-1] < prices[i])? (prices[i]-low[i-1]):0;
初始化：
f[0] = 0;
low[0] = prices[0];
题解：
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 int size = prices.size(); 5 if (size == 0) return 0; 6 int f[size], low[size], max = 0; 7 f[0] = 0; 8 low[0] = prices[0]; 9 for (int i = 1; i < size; i++) { 10 low[i] = (low[i-1] < prices[i])? low[i-1]:prices[i]; 11 f[i] = (low[i-1] < prices[i])? (prices[i]-low[i-1]):0; 12 if (f[i] > max) max = f[i]; 13 } 14 return max; 15 } 16 };