Leetcode：121. Best Time to Buy and Sell Stock

題目：
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.

例1：
輸入：[7，1，5，3，6，4]
輸出：5

最大。差額= 6-1 = 5（不是7-1 = 6，因爲賣價需要大於買入價格）
例2：
輸入：[7，6，4，3，1]
輸出：0

在這種情況下，沒有交易完成，即最大利潤= 0。

public int maxProfit(int[] prices) {
        if (prices.length==0) {
            return 0;
        }
        //初始化賣的價格就是第一個價格
        int soldPrice = prices[0];
        int max = 0;
        for (int i = 1; i < prices.length; i++) {
            //後面的賣出的價格大於前面的價格纔會盈利，否則虧本
            if (prices[i] >= soldPrice) {
                max = Math.max(max, prices[i]-soldPrice);
            }
            //第i個比當前的sold還小那肯定更新，因爲差值會變大
            else {
                soldPrice = prices[i];
            }

        }
        return max;
    }