# 1395.Count Number of Teams

**1395.Count Number of Teams**

难度:Medium

> n 名士兵站成一排。每个士兵都有一个 独一无二 的评分 rating 。 每 3 个士兵可以组成一个作战单位，分组规则如下： 从队伍中选出下标分别为 i、j、k 的 3 名士兵，他们的评分分别为 rating\[i]、rating\[j]、rating\[k] 作战单位需满足： rating\[i] < rating\[j] < rating\[k] 或者 rating\[i] > rating\[j] > rating\[k] ，其中 0 <= i < j < k < n 请你返回按上述条件可以组建的作战单位数量。每个士兵都可以是多个作战单位的一部分。

```
示例 1：

输入：rating = [2,5,3,4,1]
输出：3
解释：我们可以组建三个作战单位 (2,3,4)、(5,4,1)、(5,3,1) 。
示例 2：

输入：rating = [2,1,3]
输出：0
解释：根据题目条件，我们无法组建作战单位。
示例 3：

输入：rating = [1,2,3,4]
输出：4
 

提示：

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

可以建立一个索引，对每个元素，找出在其后比它大和小的数的索引，然后以每个比它大的数为基础，找出比这个大数更大的数的数目即可。递减以此类推。

```
class Solution {
public:
    int numTeams(vector<int>& rating) {
        vector<vector<int>>great,low;
        for(int i=0;i<rating.size();i++)
        {
            vector<int> tgreat, tlow;
            for(int j=i+1;j<rating.size();j++)
            {
                if(rating[j]>rating[i]) tgreat.push_back(j);
                if(rating[j]<rating[i]) tlow.push_back(j);
            }
            great.push_back(tgreat);
            low.push_back(tlow);
        }
        int res=0;
        for(auto v: great)
        {
        //    cout<<v.size()<<endl;
            if(v.size()>=2 )
            {
                for(auto i : v)
                res+=great[i].size();
            }
        }
               for(auto v: low)
        {
         //   cout<<v.size()<<endl;
            if(v.size()>=2 )
            {
                for(auto i : v)
                res+=low[i].size();
            }
        }
        return res;


    }
};
```

> 执行用时 :32 ms, 在所有 C++ 提交中击败了67.70%的用户\
> 内存消耗 :14.2 MB, 在所有 C++ 提交中击败了100.00%的用户


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