LeetCode 123: Best Time to Buy and Sell Stock III
2015-10-29 21:24
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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).
解题思路
思路一:动态规划法。以第i天为分界线,计算第 i 天之前进行一次交易的最大收益 preProfit[i],和第 i 天之后进行一次交易的最大收益 postProfit[i]。最后遍历一遍,max(preProfit[i]+postProfit[i]),(0≤i≤n−1) 就是最大收益。第 i 天之前和第 i 天之后进行一次的最大收益求法同 LeetCode 121。
代码如下:
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.size() < 2) return 0; int n = prices.size(); int preProfit , postProfit ; preProfit[0] = 0; int curMin = prices[0]; for (int i = 1; i < n; ++i) { preProfit[i] = max(preProfit[i-1], prices[i] - curMin); curMin = min(curMin, prices[i]); } postProfit[n-1] = 0; int curMax = prices[n-1]; for (int i = n - 2; i >= 0; --i) { postProfit[i] = max(postProfit[i+1], curMax - prices[i]); curMax = max(curMax, prices[i]); } int maxProfit = 0; for (int i = 0; i < n; ++i) { maxProfit = max(maxProfit, preProfit[i] + postProfit[i]); } return maxProfit; } };
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