leetcode 122. Best Time to Buy and Sell Stock II

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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

这一题，我们可以想象一条股票价值波动的曲线，想要得到收益最大，那么就需要在股票曲线每次爬升时，买于最低点，抛于最高点。

package leetcode;public class Best_Time_to_Buy_and_Sell_Stock_II_122 {public int maxProfit(int[] prices) {int n=prices.length;int profit=0;int buyPointer=0;int sellPointer=1;while (sellPointer < n) {while (sellPointer < n && prices[sellPointer] <= prices[buyPointer]) {buyPointer++;sellPointer++;}while (sellPointer + 1 < n && prices[sellPointer + 1] >= prices[sellPointer]) {sellPointer++;}if(sellPointer<n){profit += (prices[sellPointer] - prices[buyPointer]);}buyPointer = sellPointer + 1;sellPointer = sellPointer + 2;}return profit;}public static void main(String[] args) {// TODO Auto-generated method stubBest_Time_to_Buy_and_Sell_Stock_II_122 b=new Best_Time_to_Buy_and_Sell_Stock_II_122();int[] prices=new int[]{2,1};System.out.println(b.maxProfit(prices));}}

下面是大神的解法

//(a[i]-a[i-1])+(a[i-1]-a[i-2])=a[i]-a[i-2] which is the profits created by i and i-2
//so we travel from the end of the array and continually calculate the differece of i and i-1,
//we only sum those positive profits then the final results is the maximum profits

    if(prices.size()==0|| prices.size()==1) return 0;    int max_pro=0;    for(int i=prices.size()-1;i>0;i--){        if(prices[i]-prices[i-1]>0) max_pro+=prices[i]-prices[i-1];    }    return max_pro;