[LeetCode] 188.Best Time to Buy and Sell Stock IV 买卖股票的最佳时间之四
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 k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
这道题实际上是之前那道 Best Time to Buy and Sell Stock III 买股票的最佳时间之三的一般情况的推广，还是需要用动态规划Dynamic programming来解决，具体思路如下：
这里我们需要两个递推公式来分别更新两个变量local和global，我们其实可以求至少k次交易的最大利润。我们定义local[i][j]为在到达第i天时最多可进行j次交易并且最后一次交易在最后一天卖出的最大利润，此为局部最优。然后我们定义global[i][j]为在到达第i天时最多可进行j次交易的最大利润，此为全局最优。它们的递推式为：
local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)
global[i][j] = max(local[i][j], global[i - 1][j])，
其中局部最优值是比较前一天并少交易一次的全局最优加上大于0的差值，和前一天的局部最优加上差值后相比，两者之中取较大值，而全局最优比较局部最优和前一天的全局最优。
但这道题还有个坑，就是如果k的值远大于prices的天数，比如k是好几百万，而prices的天数就为若干天的话，上面的DP解法就非常的没有效率，应该直接用Best Time to Buy and Sell Stock II 买股票的最佳时间之二的方法来求解，所以实际上这道题是之前的二和三的综合体，代码如下：
class Solution {
public:
int maxProfit(int k, vector &prices) {
if (prices.empty()) return 0;
if (k >= prices.size()) return solveMaxProfit(prices);
int g[k + 1] = {0};
int l[k + 1] = {0};
for (int i = 0; i < prices.size() - 1; ++i) {
int diff = prices[i + 1] - prices[i];
for (int j = k; j >= 1; --j) {
l[j] = max(g[j - 1] + max(diff, 0), l[j] + diff);
g[j] = max(g[j], l[j]);
}
}
return g[k];
}
int solveMaxProfit(vector &prices) {
int res = 0;
for (int i = 1; i < prices.size(); ++i) {
if (prices[i] - prices[i - 1] > 0) {
res += prices[i] - prices[i - 1];
}
}
return res;
}
};
类似题目：
Best Time to Buy and Sell Stock with Cooldown
Best Time to Buy and Sell Stock III
Best Time to Buy and Sell Stock II
Best Time to Buy and Sell Stock
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