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123. Best Time to Buy and Sell Stock III
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).
分析:
1.对于任何一个 i, 将数列分成两部分 [0, i] [i , n-1];
2. left[i] 记录在i的左边最大profit , right[i] 记录在i的右边最大profit
3. 再扫一遍得到最大 left[i] + right[i]
O(3N)
//将数列分成两部分 [0, i] [i , n-1]; // left[i] , right[i] //再扫一遍得到最大 left[i] + right[i] public class Solution { public int maxProfit(int[] prices) { if(prices == null || prices.length <= 1) return 0; int N = prices.length; int left[] = new int[N]; left[0] = 0; int min = prices[0]; int right[] = new int[N]; right[N-1] = 0; int max = prices[N-1]; for(int i = 1 ; i < N; i++){ min = Math.min(min, prices[i]); left[i] = Math.max(left[i-1], prices[i] - min); } for(int i = N-2 ; i >= 0 ; i--){ max = Math.max(max, prices[i]); right[i] = Math.max(right[i+1], max - prices[i]); } int maxProfit = 0; for(int i = 0; i < N ; i++){ maxProfit = Math.max(maxProfit , left[i] + right[i]); } return maxProfit; } }
123. Best Time to Buy and Sell Stock III
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