122. Best Time to Buy and Sell Stock II.md

122. Best Time to Buy and Sell Stock II

###Question: 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.

Thinking:

• Method1:
  • 
  • profit的值是h1 + h2 + h3， 只要在在递增，就可以从最低点买入，在最高点卖出。

class Solution {
    public int maxProfit(int[] prices) {
        int result = 0;
        if(prices == null || prices.length == 0)
            return result;
        for(int i = 1; i < prices.length; i++){
            if(prices[i] > prices[i - 1]){
                result += prices[i] - prices[i - 1];
            }
        }
        return result;
    }
}

二刷

class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int result = 0;
        for(int i = 1; i < prices.length; i++){
            result += prices[i] > prices[i - 1] ? prices[i] - prices[i - 1]: 0;
        }
        return result;
    }
}

Third time

1. Method 1: Finding the dropping points.

class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int profit = 0, pre = -1;
        for(int i = 1; i < prices.length; i++){
            if(prices[i] < prices[i - 1]){
                if(pre != -1) profit += prices[i - 1] - prices[pre];
                else profit += prices[i - 1] - prices[0];
                pre = i;
            }
        }
        if(pre == -1) return prices[prices.length - 1] - prices[0];
        if(pre != prices.length - 1) profit += prices[prices.length - 1] - prices[pre];
        return profit;
    }
}

2. Use dp to solve this question.
   • s0(holding a stock): 1. rest 2. sell
   • s1(not holding any stock): 1.rest, 2. buy

class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int len = prices.length;
        int[] s0 = new int[len + 1];
        int[] s1 = new int[len + 1];
        s0[1] = -prices[0];
        for(int i = 2; i <= len; i++){
            s0[i] = Math.max(s0[i - 1], s1[i - 1] - prices[i - 1]);
            s1[i] = Math.max(s1[i - 1], s0[i - 1] + prices[i - 1]);
        }
        return s1[len];
    }
}