LeetCode-121. Best Time to Buy and Sell Stock (Java)

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

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题意

给定一个列数组，找出最大的差值，不过必须是高位减去低位的差值最大才行。本道题的特点就是：有顺序，求最大差

思路

需要多个变量，用来保存最大差值，以及买卖股票时的价格。

代码

public static int maxProfit(int[] prices) {        if(prices == null || prices.length <=0)        return 0;        int length = prices.length;        int i=1;        //临时差值        int difference = 0;        //买了第一天的股票        int buyPrices = prices[0];        //卖股票时的价格        int sellPrices = 0;        //最大差值        int maxDifference = 0;        while(i<length){        //如果该股票的价格大于买股票时的价格        if(prices[i]>buyPrices){        //求差值        difference = prices[i] - buyPrices;                sellPrices = prices[i];        //如果差值大于最大差值，则记录最大差值        if(difference > maxDifference){        maxDifference = difference;        }        }        //如果当天股票小于买股票时的价格，则买入最低时的股票        if(prices[i] < buyPrices){        buyPrices = prices[i];        }        i++;        }        return maxDifference;    }