【Best Time to Buy and Sell Stock III 】cpp

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).

代码：

class Solution {
public:int maxProfit(vector<int>& prices) {if (prices.empty()) return 0;const int len = prices.size();vector<int> l(len,0),r(len,0);// from left to rightl[0] = 0;int l_min = prices[0];for ( int i = 1; i < l.size(); ++i ){l_min = std::min(l_min, prices[i]);l[i] = std::max(l[i-1], prices[i]-l_min);}// from right to leftr[len-1] = 0;int r_max = prices[len-1];for ( int i = len-1; i >= 0; --i ){r_max = std::max(r_max, prices[i]);r[i] = std::max(r[i-1], r_max-prices[i]);}// travseral the best two timesint max_profit = 0;for ( int i = 0; i < prices.size(); ++i ){    max_profit = std::max(max_profit, l[i]+r[i]);}return max_profit;}
};

tips：

此题说最多交易两次，求最大获利。

直觉的想法就是，把整个时间段分割成两部分（共有prices.size()种分类方法）；分好后分别求两部分各自的最大值；这种算法是O(n²)时间复杂度的。

模仿之前求过的largest rectangle in histogram这道题的思路，能否利用dp思想，把算过的中间结果都存起来，把时间复杂度降低到O(n)。

想到这个思路就比较明确了：

1. 从左向右走一遍，l[i]存放0~i最多交易1次获利最大的值

2. 从右向左走一遍，r[i]存放i~prices.size()-1最多交易一次获利的最大值

3. 遍历数组l和数组r，通过遍历每种分割情况下的最大获利，并最终获得最终的最大获利值。

这里还有个细节可能会产生疑义：如果以第i天作为分割点，那么这第i天是算到前半截还是后半截呢？

这里分两种情况：

1. 如果“前半截的最大利润”和“后半截的最大利润”只有一截涉及到了第i天，显然l[i]+r[i]这个算法是没问题的

2. 如果“前”、“后”两截都涉及到了第i天呢？这时候有两种理解方法：

　　2.1 前后交易两次：前半截的某一天买入，第i天卖了，挣一笔；第i天卖完又买入了，到后面的某一天又卖了，挣第二笔。两笔加起来最大。

　　2.2 前后交易一次：前半截的某一天买入，第i天虽然卖了获利最大，但是不卖，留着；等到后面的某一天发现获利最大，直接挣一笔最大的，同样获利最大。

因此，无论按照哪种理解方法，l[i]+r[i]都是合理的，不会因为第i天作为分割点而产生影响。

=====================================================

第二次过这道题，思路上有个地方没有理清（红字），在当天可以选择交易或者不交易，如果不交易那么等于左边（或右边）相邻元素的值。

class Solution {
public:int maxProfit(vector<int>& prices) {if ( prices.empty() ) return 0;vector<int> l(prices.size(), 0);vector<int> r(prices.size(), 0);// lint minPrices = prices[0];for ( int i=1; i<prices.size(); ++i ){minPrices = min(prices[i], minPrices);l[i] = max(l[i-1], prices[i]-minPrices);}// rint maxPrices = prices[prices.size()-1];for ( int i=prices.size()-2; i>=0; --i ){maxPrices = max(maxPrices, prices[i]);r[i] = max(r[i+1], maxPrices - prices[i]);}// l to rint ret = 0;for ( int i=0; i<prices.size(); ++i ){ret = max(ret, l[i]+r[i]);}return ret;}
};

转载于:https://www.cnblogs.com/xbf9xbf/p/4546359.html

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