Java描述 LeetCode，121. Best Time to Buy and Sell Stock（买卖股票的时机I）

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

1 <= prices.length <= 105
0 <= prices[i] <= 104

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock

• idea1：首先想到的就是暴力解法，两重遍历取差值最大
• idea2：题目说有一天买入，再找另外一天卖出，我们可以假设第i天卖出，那么只要知道i-1天的最小值，就知道第i天卖出去的收入，这样边遍历，边更新最大值就可以了。
• idea3：利用动态规划，dp[i]表示前i天（包括i）的最小值，这样边遍历，边计算当天的收入，一边更新最大值就可以了。

我想到的是这个思路，原以为这个思路也是动态规划，但是好像写下来也不是，就是普通思路。我的代码如下，已经成功。

public static int maxProfit(int[] prices) {
    int n = prices.length;
    int min = prices[0];
    int max = 0;
    // i范围[1,n]
    int[] dp = new int[n + 1]; // 前i-1天（包括i-1）买入，第i天卖出获得的收入
    for (int i = 2; i < n + 1; i++) {
        if (min > prices[i - 2]) {
            min = prices[i - 2];
        }
        dp[i] = prices[i - 1] - min;
        if (max < dp[i]) {
            max = dp[i];
        }
    }
    System.out.println(Arrays.toString(dp));
    return max;
}

我这里是完全严格按照题目的意思，计算的前i-1天的最小值，和力扣答案里面包含当天当卖的情况不一样。

public static int maxProfit2(int[] prices) {
    int n = prices.length;
    int max = 0;
    // i范围[1,n]
    int[] dp = new int[n + 1]; // 前i天（包括i）天的历史最低值
    dp[1] = prices[0];
    for (int i = 2; i < n + 1; i++) {
        dp[i] = Math.min(dp[i - 1], prices[i - 1]);
        int cur = prices[i - 1] - dp[i];
        max = Math.max(cur, max);
    }
    System.out.println(Arrays.toString(dp));
    return max;
}

// 滚动数组的情况
public static int maxProfit3(int[] prices) {
    int n = prices.length;
    int max = 0;
    // i范围[1,n]
    int min = prices[0];
    for (int i = 2; i < n + 1; i++) {
        min = Math.min(min, prices[i - 1]);
        int cur = prices[i - 1] - min;
        max = Math.max(cur, max);
    }
    return max;
}

用的是动态规划的思想，但是很明显他算上了当天购入当天卖的情况，虽然不影响最后的结果，因为当天买，当天卖利润是0，只要结果里面有比0大的情况，这个就会被覆盖。