[leetcode]Best Time to Buy and Sell Stock II

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). 

算法思路：

计算出每个增区间的差值。求总和。

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length < 2) return 0;
        int pre = 0,  post = 1,length = prices.length;
        int buy = 0,profit = 0;
        for(int i = 0; i < prices.length; i++,pre = i - 1, post = i + 1){
            if(post < length && prices[pre] >= prices[i] && prices[post] > prices[i]){
                buy = prices[i];
            }
            if( prices[i] > prices[pre] && (post != length ? prices[i] >= prices[post] : true) ){
                profit += prices[i] - buy;
            }
        }
        return profit;
    }
}

 

第二遍优化：

上文代码是确定增区间的起始点，事实上不必确定起始点，只要关心相邻两点是否增长即可。

即：prices = {1，2，3，4，2，1}

前文做法是确定了增区间为1 ~ 4，因此受益为 4 - 1 = 3

本文做法是只与前一天作比较，如果有收益，则做交易，例如该例：

从第二个下标开始遍历，发现2 > 1，则做一次交易，收益为 2 - 1 = 1

往后遍历发现3 > 2 ，再做一次交易，3 - 2 = 1

如此往复

总收益为：（2 - 1） + （3 - 2） + （4 - 3） = 3

代码相对而言就简单多了

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