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[leetcode]123. Best Time to Buy and Sell Stock III 最佳炒股時機之三

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 (i.e., you must sell the stock before you buy again).

Example 1:

 

題目

和之前一樣,這次你至多能買賣兩次。

 

思路

1. Split the array into two parts, one trade each.

2. Say that f(i) stands for max profit in [0,i] , g(i) stands for max profix in [i, n-1] , then update and finally return max(f(i) + g(i))

 

程式碼

 1 // Best Time to Buy and Sell Stock III
2 // 時間複雜度O(n),空間複雜度O(n) 3 public class Solution { 4 public int maxProfit(int[] prices) { 5 if (prices.length < 2) return 0; 6 7 final int n = prices.length; 8 int[] f = new int[n]; 9 int[] g = new int[n]; 10 11 for (int i = 1, valley = prices[0]; i < n; ++i) {
12 valley = Math.min(valley, prices[i]); 13 f[i] = Math.max(f[i - 1], prices[i] - valley); 14 } 15 16 for (int i = n - 2, peak = prices[n - 1]; i >= 0; --i) { 17 peak = Math.max(peak, prices[i]); 18 g[i] = Math.max(g[i], peak - prices[i]); 19 } 20 21 int max_profit = 0; 22 for (int i = 0; i < n; ++i) 23 max_profit = Math.max(max_profit, f[i] + g[i]); 24 25 return max_profit; 26 } 27 }