1. Topic
n soldiers stand in a row. Each soldier has a unique score rating.
Every three soldiers to form a combat unit, grouping rules are as follows:
- Selected index from the team were three soldiers i, j, k, and their scores were
rating[i]、rating[j]、rating[k] - Combat units must meet:
rating[i] < rating[j] < rating[k] 或者 rating[i] > rating[j] > rating[k]where0 <= i < j < k < n
Please return the above conditions can be set up according to the number of combat units. Each soldier can be part of multiple combat units.
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
Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/count-number-of-teams
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.
2. Problem Solving
2.1 brute force solution
- A small amount of data, three-layer direct circulation
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
2.2 brute force optimization
- Found about each position of the number of large and small than
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
