LeetCode:Best Time to Buy and Sell Stock III
2017-10-28 15:20
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题目:
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).
思路:
这是上一篇博客的题目的类似题,不同的是上次的只允许交易一次,这次最多允许交易两次。上一篇博客的思路:从price[0]开始,找出当前最小的价格,然后得出当前最大的利润,不断更新最小价格和最大利润,直到遍历完price,从而得到最大利润。
这次依然用这个思路,但是把这两次交易分为first和second,first[i]代表从第1天到第i+1天为止交易一次得到的最大利润,显然first[0] (第一天买进,第一天卖出)为0。second[i]表示从第i+1天到最后一天交易一次得到的最大利润,显然second[size-1](最后一天买进,最后一天卖出)为0。first[i]和second[i]两次交易加起来刚好是第一天到最后一天交易两次所得的利润。(i=0或i=size-1等价于只交易一次)
要想得到交易两次的最大利润,只需要求出first[i]+second[i]的最大值。
代码:
class Solution { public: int maxProfit(vector<int>& prices) { int size = prices.size(); if (size == 0) return 0; vector<int> first(size); vector<int> second(size); int minPrice = prices[0]; first[0] = 0; for (int i = 1; i < size; i++) { minPrice = min(minPrice, prices[i]); first[i] = max(first[i-1], prices[i] - minPrice); } int maxPrice = prices[size-1]; second[size-1] = 0; for (int i = size-2; i >= 0; i--) { maxPrice = max(maxPrice, prices[i]); second[i] = max(second[i+1], maxPrice - prices[i]); } int maxPro = 0; for (int i = 0; i < size; i++) { maxPro = max(maxPro, first[i] + second[i]); } return maxPro; } };
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