Description
Given an array of integers arr, and three integers a, b and c. You need to find the number of good triplets.
A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:
- 0 <= i < j < k < arr.length
- |arr[i] - arr[j]| <= a
- |arr[j] - arr[k]| <= b
- |arr[i] - arr[k]| <= c
Where |x| denotes the absolute value of x.
Return the number of good triplets.
Example 1:
Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
Output: 4
Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].
Example 2:
Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1
Output: 0
Explanation: No triplet satisfies all conditions.
Constraints:
- 3 <= arr.length <= 100
- 0 <= arr[i] <= 1000
- 0 <= a, b, c <= 1000
分析
题目的意思是：给你一个数组，求出满足上述条件的三元组；思路很直接，需要用到三个循环，根据条件判断就行了。我以为还有更好的解法，发现别人的解法也跟我类似。
代码
class Solution:
def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
n=len(arr)
cnt=0
for i in range(n):
for j in range(i+1,n):
t1=abs(arr[i]-arr[j])
if(t1<=a):
for k in range(j+1,n):
t2=abs(arr[j]-arr[k])
t3=abs(arr[i]-arr[k])
if(t2<=b and t3<=c):
cnt+=1
return cnt
参考文献
[LeetCode] python with for loops
