【LeetCode笔记】Candy

Something wrong with my solution as describing below.

I got a wrong answer: Input: [1,2,4,4,3] Output: 10 Expected: 9

my output is 1+2+3+3+1 = 10. and I think OJ's is 1+2+3+2+1 = 9

how can the third child has more candies than the fourth?

But the requirement is "Children with a higher rating get more candies than their neighbors",

I don't want to deal with my code anymore, because I think this question is not well defined.

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I found a nice code in the discussion:
ref: https://oj.leetcode.com/discuss/76/does-anyone-have-a-better-idea
Here is another wonderful solution from old discuss by hawk. Thanks to hawk!

The solution is O(N) time complexity and constant memory complexity. What's more, the solution only need to go thru the ratings array once!

Reading the code with full comment to help understand the algorithm.

int candy(vector<int> &ratings) {

// Note: The Solution object is instantiated only once and is reused by each test case.

int nCandyCnt = 0;///Total candies

int nSeqLen = 0; /// Continuous ratings descending sequence length

int nPreCanCnt = 1; /// Previous child's candy count

int nMaxCntInSeq = nPreCanCnt;

if(ratings.begin() != ratings.end())

{

nCandyCnt++;//Counting the first child's candy.

for(vector<int>::iterator i = ratings.begin()+1; i!= ratings.end(); i++)

{

// if r[k]>r[k+1]>r[k+2]...>r[k+n],r[k+n]<=r[k+n+1],

// r[i] needs n-(i-k)+(Pre's) candies(k<i<k+n)

// But if possible, we can allocate one candy to the child,

// and with the sequence extends, add the child's candy by one

// until the child's candy reaches that of the prev's.

// Then increase the pre's candy as well.

// if r[k] < r[k+1], r[k+1] needs one more candy than r[k]

//

if(*i < *(i-1))

{

//Now we are in a sequence

nSeqLen++;

if(nMaxCntInSeq == nSeqLen)

{

//The first child in the sequence has the same candy as the prev

//The prev should be included in the sequence.

nSeqLen++;

}

nCandyCnt+= nSeqLen;

nPreCanCnt = 1;

}

else

{

if(*i > *(i-1))

{

nPreCanCnt++;

}

else

{

nPreCanCnt = 1;

}

nCandyCnt += nPreCanCnt;

nSeqLen = 0;

nMaxCntInSeq = nPreCanCnt;

}

}

}

return nCandyCnt;

}