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.
这道题比较笨的方式是O(n^2)，全都比一遍，这里做了一点小小的优化，就是如果一个数比它的下一个大，那这个数肯定不会是买入点。

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    var profit = 0;
    var days = prices.length;    
    for (var i = 0; i < days; i++) {
        if (prices[i]>prices[i+1]) {
            continue;
        }
        for (var j = i+1; j<days;j++) {
            var tamp = prices[j]-prices[i];
            if (tamp>profit)
                profit = tamp;
        }
    }
    return profit;
};

还有一种办法，只走一遍，在走的过程中把到目前为止最小的记录下来，并用当前的减去最小的，得到的值中最大的就是最大利润。

var maxProfit = function(prices) {
    // var profit = 0;
    // var days = prices.length;

        
    // for (var i = 0; i < days; i++) {
    //     if (prices[i]>prices[i+1]) {
    //         continue;
    //     }
    //     for (var j = i+1; j<days;j++) {
    //         var tamp = prices[j]-prices[i];
    //         if (tamp>profit)
    //             profit = tamp;
    //     }
    // }
    // return profit;
    var minprice = prices[0];
    var maxprofit = 0;
    var num = prices.length;
    for (var i = 0; i<num; i++){
        if (prices[i]<minprice)
            minprice = prices[i];
        if ((prices[i]-minprice)>maxprofit) {
            maxprofit = prices[i]-minprice;
        }
    }
    return maxprofit;
};