[LeetCode] 188.Best Time to Buy and Sell Stock IV 買(mǎi)賣(mài)股票的最佳時(shí)間之四

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.

這道題實(shí)際上是之前那道 Best Time to Buy and Sell Stock III 買(mǎi)股票的最佳時(shí)間之三的一般情況的推廣，還是需要用動(dòng)態(tài)規(guī)劃Dynamic programming來(lái)解決，具體思路如下：

這里我們需要兩個(gè)遞推公式來(lái)分別更新兩個(gè)變量local和global，我們其實(shí)可以求至少k次交易的最大利潤(rùn)。我們定義local[i][j]為在到達(dá)第i天時(shí)最多可進(jìn)行j次交易并且最后一次交易在最后一天賣(mài)出的最大利潤(rùn)，此為局部最優(yōu)。然后我們定義global[i][j]為在到達(dá)第i天時(shí)最多可進(jìn)行j次交易的最大利潤(rùn)，此為全局最優(yōu)。它們的遞推式為：

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])，

其中局部最優(yōu)值是比較前一天并少交易一次的全局最優(yōu)加上大于0的差值，和前一天的局部最優(yōu)加上差值后相比，兩者之中取較大值，而全局最優(yōu)比較局部最優(yōu)和前一天的全局最優(yōu)。

但這道題還有個(gè)坑，就是如果k的值遠(yuǎn)大于prices的天數(shù)，比如k是好幾百萬(wàn)，而prices的天數(shù)就為若干天的話，上面的DP解法就非常的沒(méi)有效率，應(yīng)該直接用Best Time to Buy and Sell Stock II 買(mǎi)股票的最佳時(shí)間之二的方法來(lái)求解，所以實(shí)際上這道題是之前的二和三的綜合體，代碼如下：

class Solution {
public:
    int maxProfit(int k, vector<int> &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<int> &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;
    }
};

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