Best time to buy and buy stock III

Best time to buy and buy stock III

Say you have an array for whichITh element is the price of a given stock on dayI.

Design an algorithm to find the maximum profit. You may complete at mostTwoTransactions.

 

C ++ version code:

class Solution {public:    int maxProfit(vector<int> &prices) {        int profit = 0, n = prices.size();        if( n==0 ) {            return 0;        }        int l[n], r[n];        memset(l, 0, sizeof(int)*n);        memset(r, 0, sizeof(int)*n);        int min = prices[0];        for(int i=1; i<n; i++) {            l[i] = prices[i] - min > l[i-1] ? prices[i] - min : l[i-1];            min = prices[i] < min ? prices[i] : min;        }        int max = prices[n-1];        for(int i=n-2; i>=0; i--) {            r[i] = max - prices[i] > r[i+1] ? max - prices[i] : r[i+1];            max = prices[i] > max ? prices[i] : max;        }        for (int i=0; i<n ;i++) {            profit = l[i] + r[i] > profit ? l[i] + r[i] : profit;        }        return profit;    }}; 

Java version code:

public class Solution {    public int maxProfit(int[] prices) {     if(prices == null || prices.length == 0) {         return 0;     }     int n = prices.length;     int[] left = new int[n];     int[] right = new int[n];     int min = prices[0];     for(int i=1; i<n; i++) {         left[i] = left[i - 1] > prices[i] - min ? left[i - 1] : prices[i] - min;           min = min < prices[i] ? min : prices[i];     }     int max = prices[n-1];     for(int i=n-2; i>0; i--) {         right[i] = right[i + 1] > max - prices[i] ? right[i + 1] : max - prices[i];          max = max > prices[i] ? max : prices[i];     }     int profit = 0;     for(int i=0; i<n; i++) {         profit = profit > left[i] + right[i] ? profit : left[i] + right[i];     }     return profit;    }}

　　

Best time to buy and buy stock III