请编写一个函数，接收两个参数：一个是数字列表nums，一个是数字n。返回列表中任意两个数字的差值刚好是n的所有组合的个数。
示例：
输入：(nums=[1, 1, 5, 6, 9, 16, 27], n=4)，输出3。因为组合情况可以是：(1,5), (1,5), (5,9)题目难度：简单
题目来源：codewars
def int_diff(nums: list, n: int) -> int:
pass
assert int_diff([1, 1, 5, 6, 9, 16, 27], 4) == 3
assert int_diff([1, 1, 3, 3], 2) == 4
def int_diff(nums: list, n: int):
count=0
for i in range(0,len(nums)):
for j in range(0,len(nums)):
if nums[i]-nums[j]==n:
count+=1
print(count)
return count
@jisonglin 同学太优秀了～～
考虑到题解的交流和分享，帮忙优化一下题解的格式哈，写成markdown格式的。
操作方法：将代码块的前后使用一对 ``` 符号包裹起来。
例如你的题解改成：
def int_diff(nums: list, n: int):
count=0
for i in range(0,len(nums)):
for j in range(0,len(nums)):
if nums[i]-nums[j]==n:
count += 1
return count
def int_diff(nums: list, n: int) -> tuple:
count = 0
result = list()
for i in range(len(nums) - 1):
for j in range(i + 1, len(nums)):
# 计算绝对值
if abs(nums[i] - nums[j]) == n:
count += 1
result.append((nums[i], nums[j]))
return count, result
def int_diff(nums: list, n: int) -> int:
# 定义匹配的组合个数
group_count = 0
# 第一层 for 循环
for i in range(len(nums)):
# 取第一个数
num1 = nums[i]
# 第二层 for 循环，以 i+1 开始
for j in range(i + 1, len(nums)):
# 取第二个数
num2 = nums[j]
# 判断两个数的差值是否为 n
if abs(num2 - num1) == n:
# 是则匹配的组合个数 +1
group_count = group_count + 1
# 返回匹配的组合个数
return group_count
assert int_diff([1, 1, 5, 6, 9, 16, 27], 4) == 3
assert int_diff([1, 1, 3, 3], 2) == 4
/**
* 请编写一个函数，接收两个参数：一个是数字列表nums，一个是数字n。返回列表中任意两个数字的差值刚好是n的所有组合的个数。
* 示例：
* 输入：(nums=[1, 1, 5, 6, 9, 16, 27], n=4)，输出3。因为组合情况可以是：(1,5), (1,5), (5,9)
* @param arrayList
* @param integer
* @return
*/
public Integer IntDiff(ArrayList<Integer> arrayList,Integer integer){
int result =0;
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i; j <arrayList.size() ; j++) {
if (arrayList.get(j) - arrayList.get(i) == integer){
result++;
}
}
}
return result;
}
@Test
public void TestIntDiff(){
ArrayList<Integer> integers = new ArrayList<>();
integers.add(1);
integers.add(1);
integers.add(5);
integers.add(6);
integers.add(9);
integers.add(16);
integers.add(27);
assert IntDiff(integers, 4) == 3;
ArrayList<Integer> integers1 = new ArrayList<>();
integers1.add(1);
integers1.add(1);
integers1.add(3);
integers1.add(3);
assert IntDiff(integers1, 2) ==4;
}
def int_diff(nums: list, n: int) → int:
list1=
a=len(nums)
for i in range(0,a):
for j in range(0,a):
if nums[j]-nums[i]==n:
list1.append((nums[i],nums[j]))
c=len(list1)
return c
import itertools
def int_diff(nums: list, n: int) -> list:
pairs = list(itertools.combinations(nums, 2))
valid = [(x, y) for x, y in pairs if abs(x - y) == n]
return len(valid)
解题思路：
def int_diff(nums: list, n: int) -> int:
return sum([nums.count(num + n) for num in nums if (num + n ) in nums])
assert int_diff([1, 1, 5, 6, 9, 16, 27], 4) == 3
assert int_diff([1, 1, 3, 3], 2) == 4
老师，请问写成下面这样： 为什么符合要求的数据会被打印两遍呢？
def solution(nums: list,n : int)-> int :
m=0
#b=
for i in range(0, len(nums)-1):
for j in range(i,len(nums)):
if nums[j]-nums[i]==n:
#b.append(nums[i])
#b.append(nums[j])
print(nums[i],nums[j]) 为什么被打印了两遍呢
m += 1
return m
if name == ‘main’:
a=[1, 1, 5, 6, 9, 16, 27]
n=4
solution(a,4)
print(solution(a, 4))
以下是运行结果：
1 5
1 5
5 9
1 5
1 5
5 9
3
def get_num(nums:list, n:int)->int:
sum1 = 0
for i in nums:
if i+n in nums[i:]:
sum1+= nums.count(i+n)
return sum1def int_diff(nums: list, n: int) -> int:
count = 0
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if abs(nums[j] - nums[i]) == n:
count+=1
return count
import itertools,math
def int_diff(nums:list,n:int) -> int:
return len([list(itertools.combinations(nums,2))[i] for i in range(len(list(itertools.combinations(nums,2)))) if int(math.fabs(list(itertools.combinations(nums,2))[i][0]-list(itertools.combinations(nums,2))[i][1]))== n])
assert int_diff([1,1,5,6,9,16,27],4)==3
assert int_diff([1,1,3,3],2)==4
def int_diff(nums: list, n: int) -> int:
count = 0
# result = []
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if abs(nums[i] - nums[j]) == n:
count += 1
# result.append((nums[i],nums[j]))
# print(result)
return count
assert int_diff([1, 1, 5, 6, 9, 16, 27], 4) == 3
assert int_diff([1, 1, 3, 3], 2) == 4