题目：

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). 

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题解：

买卖股票，可以多次transactions。使用Greedy贪婪法。假如每天的股票价值比前一天高，则差值可以计算入总结果中。原理是类似1,2,3,4,5，除第一天外每天先卖后买的收益其实等于局部最小值1和局部最大值5之间的收益。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {    public int maxProfit(int[] prices) {   //greedy        if(prices == null || prices.length == 0)            return 0;        int result = 0;                    for(int i = 0; i < prices.length; i++){            if(i > 0 && prices[i] - prices[i - 1] > 0)                result += prices[i] - prices[i - 1];        }                return result;    }}

 

Update:

public class Solution {    public int maxProfit(int[] prices) {        if(prices == null || prices.length == 0)            return 0;        int maxProfit = 0;                for(int i = 0; i < prices.length; i++)             if(i > 0 && prices[i] - prices[i - 1] > 0)                maxProfit += prices[i] - prices[i - 1];                return maxProfit;    }}

 

那么问题来了，为什么用Greedy这样算的结果会是最优？ 因为只有上升序列会提供profit，而在一个上升序列里比如1,2,3,4,5，最后的profit等于 p[last] - p[first]， 同时也等于所有p[i] - p[i-1]的和。discussion发现有篇写得很好，收在reference里了。

 

二刷:

和一刷方法一样。使用逐步累积的profit来求得最后总的最大profit。

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {    public int maxProfit(int[] prices) {        if (prices == null) return 0;        int max = 0;        for (int i = 1; i < prices.length; i++) {            int profit = prices[i] - prices[i - 1];            if (profit > 0) max += profit;        }        return max;    }}

 

Reference：

https://leetcode.com/discuss/6198/why-buying-day-and-selling-day-day-day-gauruntees-optmiality