题目
You are given an array prices
where prices[i]
is the price of a given stock on the ith
day.
Find the maximum profit you can achieve. 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 simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [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: prices = [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: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e., max profit = 0.
Constraints:
1 <= prices.length <= 3 * 104
0 <= prices[i] <= 104
思路
不限制交易次数,那么只要有利可图(数组靠后的价格比前一个价格要高)就可以交易,
在一个非decreasing数组上,最大值和最小值的差,等于相邻两个数字差值的和。
代码
class Solution {
public int maxProfit(int[] prices) {
int checkIdx = 1;
int result = 0;
while(checkIdx < prices.length) {
if (prices[checkIdx] >= prices[checkIdx - 1]) {
result += (prices[checkIdx] - prices[checkIdx -1]);
}
checkIdx++;
}
return result;
}
}
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