123-best time to Buy and Sell Stock III

Topic:

Say you has an array for which the i-th element is the price of a given-stock on day I.

Design an algorithm to find the maximum profit. You are in most of the transactions.

Note:
Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).

Ideas:

1. The transaction is similar to the practice of trading once, simply split the array into two arrays, and then calculate the maximum value of two array transactions once, then add;

2. Assuming that the array is split at index i, place the maximum value of the array element between 0-i into List1, and place the maximum value of the array element between the i-n into List2

3. Traverse I to determine the value of I, to obtain the maximum benefit

V1 version
public class Solution {public int maxprofit (int[] prices) {if (prices==null| |        Prices.length<=0) return 0;                       int len=prices.length;         Array length list<integer> dp_pre=new arraylist<integer> ();       0-i Sub-array list<integer> dp_post=new arraylist<integer> ();        I-n the word group int min_pre=integer.max_value;        int max_pre=0;            for (int i=0;i<len;i++) {if (prices[i]<min_pre) min_pre=prices[i];            if (prices[i]-min_pre>max_pre) Max_pre=prices[i]-min_pre;        Dp_pre.add (Max_pre); }//Fill dp_pre int max_post=p        RICES[LEN-1];        int max_pro=0;            for (int i=len-1;i>=0;i--) {if (prices[i]>max_post) max_post=prices[i]; if (Max_post-prices[i]>max_pro) max_pro=max_post-pRices[i];        Dp2.add (Max_pro);        }//fill dp_post int max=0; for (int i=1;i<len;i++) {if (Dp1.get (i) +dp2.get (len-1-i) >max) max=dp1.get (i) +dp2.get (len        -1-I);     } return max; }}

V2 version---the same idea, relatively concise
public class Solution {public    int maxprofit (int[] prices) {        if (len<=0)            return 0;        int maxpro1=0;        int min1=prices[0];        List<integer> mp=new arraylist<integer> ();        Mp.add (MaxPro1);        for (int i=1;i<len;i++) {            if (prices[i]-min1>maxpro1)                maxpro1=prices[i]-min1;            if (prices[i]<min1)                 min1=prices[i];            Mp.add (MaxPro1);        }        int maxpro2=0;        int max2=prices[len-1];        for (int i=len-2;i>=0;i--) {            if (max2-prices[i]+mp.get (i) >maxpro2) {                Maxpro2=max2-prices[i]+mp.get ( i);            }            if (PRICES[I]>MAX2)                max2=prices[i];        }        return MaxPro2;}         }

　　

123-best time to Buy and Sell Stock III