【LeetCode】122. Best Time to Buy and Sell Stock II

Difficulty:easy

 More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

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.

Intuition

Normal Method: Find the peak and valley

Clever Method: If there is profit between two days, make a deal.

Solution

    //Normal Method: Find the peak and valley
    public int maxProfit1(int[] prices) {
        if(prices==null || prices.length<=1)
            return 0;
        int i=0,j=0;
        int profit=0;
        while(i<prices.length && j<prices.length){
            while(i+1<prices.length && prices[i+1]<prices[i])
                i++;
            j=i+1;
            if(j==prices.length)
                break;
            else{
                while(j+1<prices.length && prices[j+1]>prices[j])
                    j++;
            }
            profit+=prices[j]-prices[i];
            i=j+1; //don't forget this line
        }
        return profit;
    }
    
    //Clever Method
     public int maxProfit(int[] prices) {
         if(prices==null || prices.length<=1)
            return 0;
         int profit=0;
         for(int i=0; i<prices.length-1; i++){
             if(prices[i+1]>prices[i])
                 profit+=prices[i+1]-prices[i];
         }
         return profit;
     }

　　

Complexity

Time complexity : O(n)

Space complexity : O(1)

 

 More:【目录】LeetCode Java实现