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 (i.e., buy one and sell one share of the stock multiple times).
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 1:
Input: [7,1,5,3,6,4] Output: 7 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Example 2:
Input: [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
题目描述：
一次股票交易包含买入和卖出，多个交易之间不能交叉进行。
解题思路：
对于 [a, b, c, d]，如果有 a <= b <= c <= d ，那么最大收益为 d - a。而 d - a = (d - c) + (c - b) + (b - a) ，因此当访问到一个 prices[i] 且 prices[i] - prices[i-1] > 0，那么就把 prices[i] - prices[i-1] 添加到收益中，从而在局部最优的情况下也保证全局最优。
具体解题思路及代码如下：
package Leetcode_Github;
public class GreedyThought_MaxProfit_122_1112 {
public int MaxProfit(int[] prices) {
int profit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] - prices[i - 1] > 0) {
profit += prices[i] - prices[i - 1];
}
}
return profit;
}
public static void main(String[] args) {
int[] prices = {1, 2, 3, 4, 5};
GreedyThought_MaxProfit_122_1112 test = new GreedyThought_MaxProfit_122_1112();
int result = test.MaxProfit(prices);
System.out.println(result);
}
}