问题描述
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.
解决思路
其实不用想复杂，就是只要第后一天比前一天价格高，就卖出，保证利润一直都是正的，最后相加就是最大利润了。
代码
package leetcode;
public class BestTimetoBuyandSellStockII {
public int maxProfit(int[] prices) {
int max = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] - prices[i - 1] > 0){
max += prices[i] - prices[i - 1];
}
}
return max;
}
}
