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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

拿到这个题，没思路，看了答案之后，有了下面的思路：

1、把这些数字想象成折线图，初始化数组第一个数为最小值，最大差为0。

2、折线上升阶段（prices[i-1] < prices[i]），用prices[i]减去之遍历过的最小值，得到一个差，判断是不是比之前的最大差大。

3、折线下降阶段，prices[i] 跟之前的最小元素对比，小则替换。

代码如下：

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

在Top Solution里还有一个解决方案，用的寻找最大差subarray的思想。代码如下：

public int maxProfit(int[] prices) {        int maxCur = 0, maxSoFar = 0;        for(int i = 1; i < prices.length; i++) {            maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);            maxSoFar = Math.max(maxCur, maxSoFar);        }        return maxSoFar;}