LeetCode题解系列--123. Best Time to Buy and Sell Stock III

描述

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

此题是121和122的续集，我这里使用了一个比较容易理解的方法，虽然空间开销比较大，但是其实很容易优化成空间复杂度O（1）的方法，不过为了容易理解，先讲述空间开销比较大的方法。

算法如下：

我的算法需要新开四个数组：

第一个数组：表示第i天前的股票最低价格

// min price before day i, item 0 is useless
vector<int> minPricesBeforeDayI(n, 0);
minPricesBeforeDayI[0] = prices[0];
for (int i = 1; i < n; ++i) {
minPricesBeforeDayI[i] = min(minPricesBeforeDayI[i - 1], prices[i - 1]);
}

第二个数组：表示前i天通过一次交易可以获得最大收益

// maximus profit gained by one transactions before day i
vector<int> maxProfitBeforeDayI(n, 0);
for (int i = 1; i < n; ++i) {
maxProfitBeforeDayI[i] = max(maxProfitSellOnDayI[i - 1], prices[i] - minPricesBeforeDayI[i]);
}

第三个数组：表示第i天后股票的最大价格

// maximum price after day i, item (n-1) is useless
vector<int> maxPriceAfterDayI(n, 0);
for (int i = n - 2; i > -1; --i) {
maxPriceAfterDayI[i] = max(maxPriceAfterDayI[i + 1], prices[i + 1]);
}

第四个数组：表示第i天后（包括第i天）通过一次交易可以获得最大的收益

// maximum profit can be gained by one transaction after day i, item (n-1) is useless
vector<int> maxProfitAfterDayI(n, 0);
for (int i = 0; i < n - 1; ++i) {
maxProfitAfterDayI[i] = max(maxProfitAfterDayI[i + 1], maxPriceAfterDayI[i] - prices[i]);
}

最后，找到一个i满足前i天的收益和第i天后的收益相加最大，即为所求：

int resultTwoTransactions = 0;
for (int i = 1; i < n; ++i) {
resultTwoTransactions = max(resultTwoTransactions, maxProfitBeforeDayI[i] + maxProfitAfterDayI[i]);
}

同时，我们需要考虑一次交易时获得最大的收益：（因为题目要求是最多两次交易）

int resultOneTransactions = 0;
for (int i = 1; i < n; ++i) {
resultOneTransactions = max(resultOneTransactions, maxProfitBeforeDayI[i]);
}

最后，两者的最大值即为答案：

return max(resultOneTransactions, resultTwoTransactions);

完整解答

class Solution {
public:
int maxProfit(vector<int>& prices) {
// for weird input
if (prices.size() <= 1) {
return 0;
}
int n = prices.size();
// min price before day i, item 0 is useless
vector<int> minPricesBeforeDayI(n, 0);
minPricesBeforeDayI[0] = prices[0];
for (int i = 1; i < n; ++i) {
minPricesBeforeDayI[i] = min(minPricesBeforeDayI[i - 1], prices[i - 1]);
}
// maximus profit gained by one transactions before day i
vector<int> maxProfitBeforeDayI(n, 0);
for (int i = 1; i < n; ++i) {
maxProfitBeforeDayI[i] = max(maxProfitSellOnDayI[i - 1], prices[i] - minPricesBeforeDayI[i]);
}
// maximum price after day i, item (n-1) is useless
vector<int> maxPriceAfterDayI(n, 0);
for (int i = n - 2; i > -1; --i) {
maxPriceAfterDayI[i] = max(maxPriceAfterDayI[i + 1], prices[i + 1]);
}
// maximum profit can be gained by one transaction after day i, item (n-1) is useless
vector<int> maxProfitAfterDayI(n, 0);
for (int i = 0; i < n - 1; ++i) {
maxProfitAfterDayI[i] = max(maxProfitAfterDayI[i + 1], maxPriceAfterDayI[i] - prices[i]);
}
int resultTwoTransactions = 0;
for (int i = 1; i < n; ++i) {
resultTwoTransactions = max(resultTwoTransactions, maxProfitBeforeDayI[i] + maxProfitAfterDayI[i]);
}
int resultOneTransactions = 0;
for (int i = 1; i < n; ++i) {
resultOneTransactions = max(resultOneTransactions, maxProfitBeforeDayI[i]);
}
return max(resultOneTransactions, resultTwoTransactions);
}
};

致谢

此篇博客中的算法由林翰老师提出