找到所有三元组 i,j,k 使得 i+j+k=n

Question

我已经对此进行了编码，但这很长：

for i in range(n + 1):
    for j in range(n + 1):
        for k in range(n + 1):
            if i + j + k == n:

有没有什么聪明的办法让它跑得更快？目前是 O(n^3)，这很可悲。

Answer 1

有几种可能的解决方案 - 实际上您不需要在所有这些中循环直到 N，最后一个数字完全免费。

请记住，三元组中的所有数字都必须是正数（否则答案是无限的）。

1. 如果不包括排列（即(1,2,3) 与(3,2,1) 是同一个三元组），从小到大：

   def iter_triplets(n):
       # This is the smallest number, can't be more than 1/3 of n
       for i in range(0, n//3 + 1):
           sum_left = n-i
           # This is the second smallest, can't be more than 1/2 of the sum_left or less than the first by definition
           for j in range(i, sum_left//2 + 1):
               yield (i, j, sum_left-j)  # Last number is calculated.
   
   
   >>> list(iter_triplets(6))  
   [(0, 0, 6), (0, 1, 5), (0, 2, 4), (0, 3, 3), (1, 1, 4), (1, 2, 3), (2, 2, 2)]
   >>> list(iter_triplets(10))
   [(0, 0, 10), (0, 1, 9), (0, 2, 8), (0, 3, 7), (0, 4, 6), (0, 5, 5), (1, 1, 8), (1, 2, 7), (1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4)]
   

2. 如果不包括排列（即(1,2,3) 与(3,2,1) 是同一个三元组），从大到小：

   import math
   def iter_triplets(n):
       # This is the biggest number, can't be less than 1/3 of n
       for i in range(n, math.ceil(n/3) - 1, -1):
           sum_left = n-i
           # This is the second biggest number, can't be less than 1/2 of the sum_left and more than first number by definition.
           # ceil to correct rounding errors.
           for j in range(min(sum_left, i), math.ceil((sum_left)/2) - 1, -1):
               yield (i, j, sum_left-j)  # Last number is calculated.
   
   
   >>> list(iter_triplets(10))
   [(10, 0, 0), (9, 1, 0), (8, 2, 0), (8, 1, 1), (7, 3, 0), (7, 2, 1), (6, 4, 0), (6, 3, 1), (6, 2, 2), (5, 5, 0), (5, 4, 1), (5, 3, 2), (4, 4, 2), (4, 3, 3)]
   

3. 如果包含排列（即(1,2,3) 与(3,2,1) 是不同三元组），从小到大：

   def iter_triplet_permutations(n):
       for i in range(0, n+1):
           sum_left = n-i
           for j in range(0, sum_left+1):
               yield (i, j, sum_left-j)
   
   >>> list(iter_triplet_permutations(5)) 
   [(0, 0, 5), (0, 1, 4), (0, 2, 3), (0, 3, 2), (0, 4, 1), (0, 5, 0), (1, 0, 4), (1, 1, 3), (1, 2, 2), (1, 3, 1), (1, 4, 0), (2, 0, 3), (2, 1, 2), (2, 2, 1), (2, 3, 0), (3, 0, 2), (3, 1, 1), (3, 2, 0), (4, 0, 1), (4, 1, 0), (5, 0, 0)]
   

Answer 2

最里面的循环似乎是多余的，因为一旦有了i 和j，k 就会免费提供：

for i in range(n + 1):
    for j in range(n + 1):
        if i+j <= n:
            print((i, j, n-i-j))

如果我们想要添加到n 的唯一数字的三元组，那么也许这可以工作：

for i in range(n + 1):
    for j in range(i, (n - i)//2 + 1):
        out.append((i, j, n-i-j))

Answer 3

有几种方法可以让代码运行得更快。

1. 使用列表推导。语法 [ReturnThing ForLoops 条件]

2. 从前一个循环到达的地方开始每个循环。这样可以避免重复

   def MakeTriplet(n):                                                                          
       return [(i,j,k) for i in range(0,n+1) for j in range(i,n+1) for k in range(j,n+1) if (i+j+k)==n]