每日两题8

买卖股票的最佳时机 III

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int INF = 0x3f3f3f3f;
        vector<vector<int>> f(n, vector<int>(3, -INF));
        auto g = f;
        g[0][0] = 0;
        f[0][0] = -prices[0];
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < 3; j++) {
                f[i][j] = max(f[i - 1][j], g[i - 1][j] - prices[i]);
                // 细节 如果不这样处理就会出现dp表第一列填表错误
                g[i][j] = g[i - 1][j];
                if (j > 0)
                    g[i][j] = max(g[i][j], f[i - 1][j - 1] + prices[i]);
            }
        }
        int ret = 0;
        for (int j = 0; j < 3; j++)
            ret = max(ret, g[n - 1][j]);
        return ret;
    }
};

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买卖股票的最佳时机 IV

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        int INF = 0x3f3f3f3f;
        k = min(n / 2, k);
        vector<vector<int>> f(n, vector<int>(k + 1, -INF));
        auto g = f;

        g[0][0] = 0;
        f[0][0] = -prices[0];

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < k + 1; j++) {
                f[i][j] = max(f[i - 1][j], g[i - 1][j] - prices[i]);
                g[i][j] = g[i - 1][j];
                if (j > 0)
                    g[i][j] = max(g[i][j], f[i - 1][j - 1] + prices[i]);
            }
        }
        int ret = 0;
        for (int j = 0; j < k + 1; j++)
            ret = max(ret, g[n - 1][j]);
        return ret;
    }
};