leetcode candy

 1 int candy(vector<int> &ratings){
 2     int size = ratings.size();
 3     vector<int> count(size, 1);
 4     if (size == 0) return 0;
 5     else if (size == 1) return 1;
 6     for (int i = 1; i < size; i++){
 7         if (ratings[i] > ratings[i - 1]){
 8             count[i] = count[i - 1] + 1;
 9         }
10     }
11     for (int i = size - 2; i >= 0; i--){
12         if (ratings[i] > ratings[i + 1] && count[i] <= count[i + 1]){
13             count[i] = count[i + 1] + 1;
14         }
15 
16     }
17     int result = 0;
18     for (int i = 0; i < size; i++){
19         result += count[i];
20     }
21     return result;
22 }

 1 int candy(vector<int> &ratings) {
 2     // Note: The Solution object is instantiated only once and is reused by each test case.
 3     int nCandyCnt = 0;///Total candies
 4     int nSeqLen = 0;  /// Continuous ratings descending sequence length
 5     int nPreCanCnt = 1; /// Previous child‘s candy count
 6     int nMaxCntInSeq = nPreCanCnt;
 7     if (ratings.begin() != ratings.end())
 8     {
 9         nCandyCnt++;//Counting the first child‘s candy.
10         for (vector<int>::iterator i = ratings.begin() + 1; i != ratings.end(); i++)
11         {
12             // if r[k]>r[k+1]>r[k+2]...>r[k+n],r[k+n]<=r[k+n+1],
13             // r[i] needs n-(i-k)+(Pre‘s) candies(k<i<k+n)
14             // But if possible, we can allocate one candy to the child,
15             // and with the sequence extends, add the child‘s candy by one
16             // until the child‘s candy reaches that of the prev‘s.
17             // Then increase the pre‘s candy as well.
18 
19             // if r[k] < r[k+1], r[k+1] needs one more candy than r[k]
20             // 
21             if (*i < *(i - 1))
22             {
23                 //Now we are in a sequence
24                 nSeqLen++;
25                 if (nMaxCntInSeq == nSeqLen)
26                 {
27                     //The first child in the sequence has the same candy as the prev
28                     //The prev should be included in the sequence.
29                     nSeqLen++;
30                 }
31                 nCandyCnt += nSeqLen;
32                 nPreCanCnt = 1;
33             }
34             else
35             {
36                 if (*i > *(i - 1))
37                 {
38                     nPreCanCnt++;
39                 }
40                 else
41                 {
42                     nPreCanCnt = 1;
43                 }
44                 nCandyCnt += nPreCanCnt;
45                 nSeqLen = 0;
46                 nMaxCntInSeq = nPreCanCnt;
47             }
48         }
49     }
50     return nCandyCnt;
51 }