leetcode-121-easy

Best Time to Buy and Sell Stock

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

思路一：最开始想到的是最简单的解法，双层循环，遍历所有可能性，但是超时了。后来想想，维护两个值，一个是当前最优解，一个是当前最小的值，每次遍历下一个数字的时候，更新这个两个值，就能求解。评论区有个解释很好

动态规划 前i天的最大收益 = max{前i-1天的最大收益，第i天的价格-前i-1天中的最小价格}

public int maxProfit(int[] prices) {
    if (prices.length < 2) return 0;

    int maxVal = prices[0] < prices[1] ? prices[1] - prices[0] : 0;
    int min = Math.min(prices[0], prices[1]);

    for (int i = 2; i < prices.length; i++) {
        if (prices[i] < min) {
            min = prices[i];
        } else {
            maxVal = Math.max(prices[i] - min, maxVal);
        }
    }

    return maxVal;
}