这题就是从左到右对递增的点进行刷点，再从右到左对递增的点进行刷点。时间复杂度和空间复杂度都为O(n)。注意如果两个相邻的点ratings一样，则不必保持candy数一样（个人认为很不合理）.

1 class Solution { 2 public: 3     int candy(vector
     
       &ratings) { 4         // IMPORTANT: Please reset any member data you declared, as 5         // the same Solution instance will be reused for each test case. 6         vector
      
        res(ratings.size(), 1); 7         for (int i = 1; i < ratings.size(); i++)  8             if (ratings[i] > ratings[i-1]) res[i] = res[i-1] + 1; 9         for (int i = ratings.size()-2; i >= 0; i--) 10             if (ratings[i] > ratings[i+1]) res[i] = max(res[i], res[i+1] + 1);11         int ans = 0;12         for (int i = 0; i < res.size(); i++) ans += res[i];13         return ans;14     }15 };
      
     

 C#

1 public class Solution { 2     public int Candy(int[] ratings) { 3         int[] res = new int[ratings.Length]; 4         for (int i = 0; i < ratings.Length; i++) res[i] = 1; 5         for (int i = 1; i < ratings.Length; i++) { 6             if (ratings[i] > ratings[i-1]) res[i] = res[i-1] + 1; 7         } 8         for (int i = ratings.Length-2; i >= 0; i--) { 9             if (ratings[i] > ratings[i+1]) res[i] = Math.Max(res[i], res[i+1] + 1);10         }11         int ans = 0;12         for (int i = 0; i < res.Length; i++) ans += res[i];13         return ans;14     }15 }