Q：
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.
A：
找到最小值出吹，然后找到之后出現(xiàn)的最大值既们，differ就是max_profit。
https://leetcode.com/articles/best-time-buy-and-sell-stock/
需要設(shè)置兩個(gè)量：1.用來存儲(chǔ)已出現(xiàn)過的min 2.如果下一個(gè)值min大惑朦，那么計(jì)算差值，并且儲(chǔ)存。one pass結(jié)束時(shí)，min為谷底值兼雄，max_profit = 頂點(diǎn)值-谷底值。
public class Solution {
public int maxProfit(int prices[]) {
int minprice = Integer.MAX_VALUE;
int maxprofit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] < minprice)
minprice = prices[i];
else if (prices[i] - minprice > maxprofit)
maxprofit = prices[i] - minprice;
}
return maxprofit;
}
}
