LeetCode動態規劃思想：Best Time to Buy and Sell Stock III

題目：

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

動態規劃的思想

如數組 [6，2，4，5]

正向掃描

f=[0，0，2，3]

逆向掃描

g=[0，3，1，0]

maxProfit = max(f[i]+g[i]);

因爲f[3] = 2 g[3] = 1;

而f[3]代表的是2，4；

g[3]代表的是4，5；

這樣就不會重疊了而且保證了是兩段裏面的最大值；

代碼：

int maxProfit4( int[] prices) {
           if (prices.length < 2)
               return 0;
           int n = prices.length ;
           int [] f = new int[n];
           int [] g = new int[n];

           for (int i = 1, valley = prices[0]; i < n; ++i) {
              valley = min(valley, prices[i]);
              f[i] = max(f[i - 1], prices[i] - valley);
          }
           for (int i = n - 2, peak = prices[n - 1]; i >= 0; --i) {
              peak = max(peak, prices[i]);
              g[i] = max(g[i], peak - prices[i]);
          }
           int max_profit = 0;
           for (int i = 0; i < n; ++i)
              max_profit = max(max_profit, f[i] + g[i]);
           return max_profit;
     }