LWC 71: 781. Rabbits in Forest

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LWC 71: 781. Rabbits in Forest

传送门：781. Rabbits in Forest

Problem:

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.

Examole:

Input: answers = [1, 1, 2]
Output: 5
Explanation:
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

Note:

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

思路：
贪心，如果出现[1, 1, 1, 1]这种情况，组合{1, 1}为同一颜色的兔子，这样情况最少[{1, 1}, {1, 1}]总共有4只兔子。再看[2, 2, 2, 2, 2]，可以组合{2, 2, 2} 和{2, 2}。所以先统计频次，后组合。

代码如下：

    public int numRabbits(int[] answers) {
        int[] map = new int[1024];
        for (int ans : answers) {
            map[ans]++;
        }
        int ans = 0;
        for (int i = 0; i < 1024; ++i) {
            if (map[i] != 0) {
                int batch = i + 1;
                int count = map[i] / batch;
                if (map[i] % batch != 0) count++;
                ans += batch * count;
            }
        }
        return ans;
    }

python 版本：

class Solution(object):
    def numRabbits(self, answers):
        """
        :type answers: List[int]
        :rtype: int
        """
        from collections import Counter
        count = Counter(answers)
        ans = 0
        for val, freq in count.items():
            ans += (freq + val) // (val + 1) * (val + 1)
        return ans