[leetcode]_Best Time to Buy and Sell Stock I II

一个系列三道题，我都不会做，google之答案。过了两道，第三道看不懂，放置，稍后继续。

一、Best Time to Buy and Sell Stock I

题目：一个数组表示一支股票的价格变换。要求只买卖一次，获得最大收益。

思路：一开始我认为是寻找最大、最小值，但由于最大值不一定总是出现在最小值的后面，因此WA。

参考思路：DP。对第i个价格，减去前i-1个价格中的最小值（保证该收益是在第i个价格卖出的最大收益），其收益与之前获得的最大收益相比。

代码：

 1 public int maxProfit(int[] prices) {
 2         int len = prices.length ;
 3         if(len < 2) return 0;
 4
 5         int min = prices[0] ;
 6         int maxProfit = 0;
 7
 8         for(int i = 1 ; i < len ; i++){
 9             int temp = prices[i] - min; //当前值减去前i-1个值的最小值
10             if(maxProfit < temp) maxProfit = temp; //更新最大收益
11             if(prices[i] < min) min = prices[i]; //看是否需要更新前i个值的min值，用于下次循环
12         }
13
14         return maxProfit;
15 }

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

二、Best Time to Buy and Sell Stock II

题目：在上一题的基础上，允许对一支股票任意次买卖，（同一时间可先卖出再马上买入），同样求最大收益。

思路：如果第i个价格大于第i-1个价格，则将此部分收益加入到最大收益中，因为可以在第i个价格处马上卖出再马上买入。

代码：

1 public int maxProfit(int[] prices) {
2         int profit = 0;
3         for(int i = 1 ; i < prices.length ; i++){
4             if(prices[i] > prices[i - 1]) profit += prices[i] - prices[i - 1];
5         }
6         return profit;
7 }

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

三、待续

转载于:https://www.cnblogs.com/glamourousGirl/p/3753849.html

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