<leetcode刷题-数组> 【动态规划】【贪心算法】买卖股票的最佳时机

#define MAX(a, b) ((a > b) ? a : b)
int maxProfit(int* prices, int pricesSize)
{
    int (*dp)[2] = (int(*)[2])malloc(sizeof(int) * pricesSize * 2);

    if ((prices == NULL) || (pricesSize < 2))
        return 0;

    // 初始条件
    dp[0][0] = 0; // 第一天没有买股票
    dp[0][1] = -prices[0]; // 第一天买了股票

    // 注意这里的初始条件day从1开始,即从第二天开始.
    for (int day = 1; day < pricesSize; day++) {
        dp[day][0] = MAX(dp[day - 1][0], dp[day - 1][1] + prices[day]);
        dp[day][1] = MAX(dp[day - 1][1], dp[day - 1][0] - prices[day]);
    }

    // 最后一天手里没有持有股票,就是最大利润
    return dp[pricesSize - 1][0];
}

int maxProfit(int* prices, int pricesSize)
{
    int profits = 0;

    if ((prices == NULL) || (pricesSize < 2))
        return 0;

    for (int i = 0; i < pricesSize - 1; i++) {
        if (prices[i] < prices[i + 1]) {
            profits += prices[i + 1] - prices[i];
        }
    }
    return profits;
}