【leetcode】1395. Count Number of Teams

题目如下：

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

• Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
• A team is valid if:  (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1).

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can‘t form any team given the conditions.

Example 3:

Input: rating = [1,2,3,4]
Output: 4

Constraints:

• n == rating.length
• 1 <= n <= 200
• 1 <= rating[i] <= 10^5

解题思路：这个题目不难。假设把rating[i]放在三个人的中间，我们只需要分别求出rating[i]左边比其大和比其小的元素的个数，以及分别求出rating[i]右边比其大和比其小的元素的个数，最后左边小的个数*右边大的个数+左边大的个数*右边小的个数，即为rating[i]放在三个人的中间时可以组成的排列数的总数。

代码如下：

class Solution(object):
    def numTeams(self, rating):
        """
        :type rating: List[int]
        :rtype: int
        """
        res = 0
        for i in range(1,len(rating)-1):
            left_small = 0
            left_great = 0
            for j in range(i):
                if rating[i] > rating[j]:
                    left_small += 1
                elif rating[i] < rating[j]:
                    left_great += 1

            right_small = 0
            right_great = 0
            for j in range(i+1,len(rating)):
                if rating[i] > rating[j]:
                    right_small += 1
                elif rating[i] < rating[j]:
                    right_great += 1

            res += left_small * right_great
            res += left_great * right_small
            #print rating[i],left_small,left_great,right_small,right_great
        return res

原文地址：https://www.cnblogs.com/seyjs/p/12631498.html