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Leetcode122-bestTimeToBuyAndSellStockII

Description

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

Example 1:

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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:
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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:
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Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Solution

Solution 1: Greedy

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class Solution {
// Greedy
public int maxProfit(int[] prices) {
int res = 0;
for (int i = 1; i < prices.length; i++)
res += Math.max(prices[i] - prices[i - 1], 0);
return res;
}
}

Solution 2: If we cannot buy and sell stocks at one day.

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class Solution {
// If we cannot buy and sell stocks at one day.
public int maxProfit(int[] prices) {
int profit = 0, i = 0;
while (i < prices.length) {
// find next local minimum
while (i < prices.length-1 && prices[i+1] <= prices[i]) i++;
int min = prices[i++]; // need increment to avoid infinite loop for "[1]"
// find next local maximum
while (i < prices.length-1 && prices[i+1] >= prices[i]) i++;
profit += i < prices.length ? prices[i++] - min : 0;
}
return profit;
}
}