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.

題意：用一個數組表示股票每天的波動價格，如果你只有一次買賣的機會，那么你能賺到的最大利潤是多少。

思路：
第一種思路，是以每個價格作為買入價格，然后再它后面找最高的賣出價格，通過兩次循環可以算出，時間復雜度是n方。

第二種思路，實際上每個價格都作為買入價格是不必要的，比如[2，3，9，20]，3和9都比2大，所以以3和9作為買入價格，肯定沒有以2獲取的利潤多。由此想到只需要知道當前最低的買入價格，如果當前價格比最低價格還低，那么更新最低價格；如果比最低價格高，那么算一下以當前價格作為賣出價格的利潤，再與最大利潤值比較。

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

    int curMin = prices[0];
    int profit = 0;
    for (int i = 1; i < prices.length; i++) {
        if (prices[i] > curMin) {
            profit = Math.max(profit, prices[i] - curMin);
        } else {
            curMin = prices[i];
        }
    }

    return profit;
}