Leetcode：121. Best Time to Buy and Sell Stock

题目：
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：
输入：[7，1，5，3，6，4]
输出：5

最大。差额= 6-1 = 5（不是7-1 = 6，因为卖价需要大于买入价格）
例2：
输入：[7，6，4，3，1]
输出：0

在这种情况下，没有交易完成，即最大利润= 0。

public int maxProfit(int[] prices) {        if (prices.length==0) {            return 0;        }        //初始化卖的价格就是第一个价格        int soldPrice = prices[0];        int max = 0;        for (int i = 1; i < prices.length; i++) {            //后面的卖出的价格大于前面的价格才会盈利，否则亏本            if (prices[i] >= soldPrice) {                max = Math.max(max, prices[i]-soldPrice);            }            //第i个比当前的sold还小那肯定更新，因为差值会变大            else {                soldPrice = prices[i];            }        }        return max;    }