【DP + 卖股票】LeetCode 188. Best Time to Buy and Sell Stock IV

最新推荐文章于 2025-04-18 22:24:21 发布

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#LeetCode 188. Best Time to Buy

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本文提供了一种解决LeetCode 188题目的方法，该题目为“最佳买卖时机IV”。文章中详细阐述了如何根据不同条件(k大于等于价格数组大小或否)选择合适的解决方案，并给出了具体的C++代码实现。

LeetCode 188. Best Time to Buy and Sell Stock IV

Solution1：我的答案
参考链接：http://www.cnblogs.com/grandyang/p/4295761.html
当k >= prices.size()时，这道题是II，否则是III

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (prices.empty()) return 0;
        if (k >= prices.size()) return other_max(prices);
        int n = prices.size(), g[n][k+1] = {0}, l[n][k+1] = {0};
        memset(g, 0, sizeof(g));
        memset(l, 0, sizeof(l));
        for (int i = 1; i < prices.size(); ++i) {
            int diff = prices[i] - prices[i - 1];
            for (int j = 1; j <= k; ++j) {
                l[i][j] = max(g[i - 1][j - 1] + max(diff, 0), l[i - 1][j] + diff);
                g[i][j] = max(l[i][j], g[i - 1][j]);
            }
        }
        return g[n - 1][k];
    }
private:
    int other_max(vector<int>& prices) {
        int n = prices.size();
        int dp[n] = {0}, res = 0;
        for (int i = 1; i < n; i++) {
            dp[i] = max(0, prices[i] - prices[i-1]);
            res += dp[i];
        }
        return res;
    }
};