[Leetcode]121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

这是第一个我自己想出的动态规划题，开心！！
思路：对于第 n 个数，假如我知道了在他之前的 n-1 个数中最小的那个数 min，并且也知道了此时的 max difference，那么
　　  答案不就是 max{ max difference, a[n] - min }吗？状态转移方程就出来了。

class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length < 1)
            return 0;    
        int min = prices[0], maxdif = 0;
        for (int i = 0; i < prices.length; i++){
            maxdif = Math.max(maxdif, prices[i] - min);
            min = Math.min(min, prices[i]);
        }
        return maxdif;
    }
}