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:
rating[i]、rating[j]、rating[k]
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
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/count-number-of-teams 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
int numTeams(vector<int>& rating) {
int i, j, k, sum = 0, n = rating.size();
for(i = 0; i < n-2; ++i)
for(j = i+1; j < n-1; ++j)
for(k = j+1; k < n; ++k)
if((rating[i] < rating[j] && rating[j] < rating[k])
||(rating[i] > rating[j] && rating[j] > rating[k]))
sum++;
return sum;
}
};
140 ms 7.6 MB
sum += lsmall*rLarge + lLarge*rsmall
class Solution {
public:
int numTeams(vector<int>& rating) {
int i, j, lsmall, lLarge, rsmall, rLarge, sum = 0, n = rating.size();
for(i = 1; i < n-1; ++i)
{
lsmall = lLarge = rsmall = rLarge = 0;
for(j = 0; j < i; ++j)
{
if(rating[j] < rating[i])
lsmall++;
if(rating[j] > rating[i])
lLarge++;
}
for(j = i+1; j < n; ++j)
{
if(rating[j] < rating[i])
rsmall++;
if(rating[j] > rating[i])
rLarge++;
}
sum += lsmall*rLarge + lLarge*rsmall;
}
return sum;
}
};
8 ms 7.3 MB