LeetCode: Best Time to Buy and Sell Stock III

思路对的，但是没弄好code，改了挺多，最后还是runtime exceed了，看了网上答案再把自己的code改了改

 1 class Solution {

 2 public:

 3     int maxProfit(vector<int> &prices) {

 4         // Start typing your C/C++ solution below

 5         // DO NOT write int main() function

 6         if (prices.size() <= 1) return 0;

 7         int size = prices.size();

 8         vector<int> left(size);

 9         vector<int> right(size);

10         int minprice = prices[0];

11         left[0] = 0;

12         for (int i = 1; i < size; i++) {

13             minprice = min(minprice, prices[i]);

14             left[i] = max(left[i-1], prices[i]-minprice);

15         }

16         int maxprice = prices[size-1];

17         right[size-1] = 0;

18         for (int i = size-2; i >= 0; i--) {

19             maxprice = max(maxprice, prices[i]);

20             right[i] = max(right[i+1], maxprice - prices[i]);

21         }

22         int maxprofit = 0;

23         for (int i = 0; i < size; i++) {

24              maxprofit = max(maxprofit, left[i]+right[i]);

25         }

26         return maxprofit;

27     }

28 };

 C#:

 1 public class Solution {

 2     public int MaxProfit(int[] prices) {

 3         int size = prices.Count();

 4         if (size < 1) return 0;

 5         int[] leftProfit = new int[size];

 6         int[] rightProfit = new int[size];

 7         int leftMinPrice = prices[0];

 8         int rightMaxPrice = prices[size-1];

 9         leftProfit[0] = 0;

10         rightProfit[size-1] = 0;

11         for (int i = 1; i < prices.Count(); i++) {

12             leftProfit[i] = Math.Max(leftProfit[i-1], prices[i] - leftMinPrice);

13             leftMinPrice = Math.Min(leftMinPrice, prices[i]);

14         }

15         for (int i = size-2; i >= 0; i--) {

16             rightProfit[i] = Math.Max(rightProfit[i+1], rightMaxPrice - prices[i]);

17             rightMaxPrice = Math.Max(rightMaxPrice, prices[i]);

18         }

19         int maxProfit = 0;

20         for (int i = 0; i < size; i++) maxProfit = Math.Max(maxProfit, leftProfit[i] + rightProfit[i]);

21         return maxProfit;

22     }

23 }

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