LeetCode - Best Time to Buy and Sell Stock II
Problem description
Say you have an array prices 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:
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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.
Constraints:
- 1 <= prices.length <= 3 * 10 ^ 4
- 0 <= prices[i] <= 10 ^ 4
Analysis
Generally, we analysis this question, and find out we need to buy at min price and sell at max price to get a max profit.
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class Solution {
public int maxProfit(int[] prices) {
if (prices.length == 0){
return 0;
}
int max = 0;
int buy = prices[0], cur = 0;
for (int i = 1;i<prices.length;i++){
if (prices[i-1] < prices[i]){
cur = Math.max(cur, prices[i] - buy);
}
else{
max += cur;
buy = prices[i];
cur = 0;
}
}
max+=cur;
return max;
}
}
However, we can get a better solution by think like this: we can improve the solution because if prices[i-1] < prices[i], the current can be divide into different transactions.
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class Solution {
public int maxProfit(int[] prices) {
int maxprofit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1])
maxprofit += prices[i] - prices[i - 1];
}
return maxprofit;
}
}