781. Rabbits in Forest 森林中的兔子

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/




In a forest, each rabbit has some color. Some subset of rabbits (possibly all of them) tell you how many other rabbits have the same color as them. Those answers are placed in an array.

Return the minimum number of rabbits that could be in the forest.


Input: answers = [1, 1, 2]
Output: 5
The two rabbits that answered "1" could both be the same color, say red.
The rabbit than answered "2" can't be red or the answers would be inconsistent.
Say the rabbit that answered "2" was blue.
Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.

Input: answers = [10, 10, 10]
Output: 11

Input: answers = []
Output: 0


  1. answers will have length at most 1000.
  2. Each answers[i] will be an integer in the range [0, 999].





If x+1 rabbits have same color, then we get x+1 rabbits who all answer x. now n rabbits answer x. If n%(x+1)==0, we need n/(x+1) groups of x+1 rabbits. If n%(x+1)!=0, we need n/(x+1) + 1 groups of x+1 rabbits. the number of groups is math.ceil(n/(x+1)) and it equals to (n+i)/(i+1) , which is more elegant.



若 n%(x+1)==0,说明我们此时只需要 n/(x+1) 组个数为x+1的兔子。

若 n%(x+1)!=0,说明我们此时只需要 n/(x+1) + 1 组个数为x+1的兔子。

那么这两种情况可以通过 ceil(n/(x+1)) 来整合,而这个值也等于 (n + x) / (x + 1).

可以理解为,对每个数字进行记数,如果其出现的次数n能整除(x+1),说明能够成一个颜色,所以x+1个兔子。如果不能整除的话,说明有n/(x+1) + 1组兔子。然后一个简化计算的方案。


class Solution(object):
    def numRabbits(self, answers):
        :type answers: List[int]
        :rtype: int
        count = collections.Counter(answers)
        print count
        return sum((count[x] + x) / (x + 1) * (x + 1) for x in count)


class Solution {
    int numRabbits(vector<int>& answers) {
        int res = 0;
        unordered_map<int, int> m;
        for (int a : answers) m[a]++;
        for (auto a : m) {
            res += (a.second + a.first) / (a.first + 1) * (a.first + 1);
        return res;


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