Leetcode 123. best time to Buy and Sell Stock III Java language

Say you had an array for which the ith element was the price of a given stock on day i.design an algorithm to find the max Imum profit. You are in most of the transactions. Note:you engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). Subscribe to see which companies asked this question.

Test instructions: Buy and sell stocks. You only have two chances to buy and sell.

Public class solution {    public int maxprofit (int[] prices )  {        if (prices==null | | &NBSP;PRICES.LENGTH&LT;2) return 0;        //o (n^2) Hey          int sum=0;        for (int  i=1;i<prices.length;i++) {            int  temp=getmax (prices,0,i) +getmax (prices,i+1,prices.length-1);             if (temp>sum) sum=temp;        }         return sum;    }     Public static int getmax (int[] prices,int left,int right) {         if (left>=prices.length) return 0;        int min=prices[left];         int sum=0;        for (int i=left+1;i <=right;i++) {            min=math.min (Min, Prices[i]);             sum=math.max (sum,prices[i]- Min);        }         System.out.println (sum);        return sum;     // }    // //first[i] From left to right, indicating to the first day to sell the maximum money can be earned     //  int[] first=new int[prices.length];    // //second[i] is traversed from right to left, Represents the maximum money that can be earned from day I to the last day. The essence is still divided into two parts calculation. It's just a separate two-scan array     // int[] second=new int[prices.length];     //  Int min=prices[0];    // for (int i=1;i<prices.length;i++) {     //     min=math.min (Min,prices[i]);    //      first[i]=math.max (first[i-1],prices[i]-min);    // }     // // for (int i=0;i<prices.length;i++) {    // //      system.out.println (First[i]);    // // }     // int max=prices[prices.length-1];    // for (int i= prices.length-2;i>=0;i--) {    //     max=math.max (max,prices[i ]);     //     second[i]=math.max (Second[i+1],max-prices[i]);     // }    // int ret=0;    //  for (int i=0;i<prices. length;i++) {    //     //system.out.println (second[i]);     //     ret=math.max (First[i]+second[i],ret);     // }        // return ret;}}

PS: Solution One is directly to the first day of the boundary, to seek the maximum benefit on both sides. However, O (n^2) is required.

Optimization: Use two arrays, traverse two times.

Leetcode 123. best time to Buy and Sell Stock III Java language