122 Best Time to Buy and Sell Stock II

122. Best Time to Buy and Sell Stock II

1. Question

Say you have an array for which thei_thelement 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.

2. Implementation

(1) Greedy

思路: 因为交易没有限制次数，所以维护两个变量，一个局部最大curMax, 一个全局最大max，如果在第i天的价格比前一天高，则买入获得利润prices[i] - prices[i - 1]，将这笔利润加到curMax，否则curMax维持不变。然后max每次都和curMax比较取较大值即可

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) {
            return 0;
        }

        int n = prices.length;
        int max = 0;
        int curMax = 0;

        for (int i = 1; i < n; i++) {
            curMax = prices[i] > prices[i - 1] ? curMax + prices[i] - prices[i - 1] : curMax;
            max = Math.max(max, curMax);
        }
        return max;
    }
}

3. Time & Space Complexity

Greedy: 时间复杂度O(n), 空间复杂度O(n)