三元组，二元组，排列组合

2024年6月9日 191次阅读 
"""
    算法题：二元组
    Given an array of integers, return indices of the two numbers such that they add up to a specific target.

    You may assume that each input would have exactly one solution, and you may not use the same element twice.
    Example:
    Given nums = [2, 7, 11, 15], target = 9,

    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].
    
    auther:Jacob
"""
class Solution:
    def permute(self,numbers,target,num):
        result = []
        permute_result = list(itertools.permutations(numbers,num))
        for i in permute_result:
            if sum(i) == target:
                result.append(i)

        return  result



if __name__ == '__main__':
    nums = [2, 7, 11, 15]
    solution = Solution()
    print(solution.permute(nums,9,2))








"""

 * 算法题：三元组
 * 给出一个有 n 个整数的数组 S，在 S 中找到三个整数 a, b, c，使得 a + b + c = 0。写一个函数找到所有满足要求的三元组。
 *
        注意事项：
       在三元组(a, b, c)，要求a <= b <= c。结果不能包含重复的三元组。

       格式：

        输入行输入一个有 n 个整数的数组 S，最后输出所有满足要求的三元组。

       样例输入

   S = [ -1，0，1，2，-1，-4 ]

       样例输出

   ( -1, 0, 1 )
   ( -1, -1, 2 )

"""
class Solution:
    """
    @param numbersbers : Give an array numbersbers of n integer
    @return : Find all unique triplets in the array which gives the sum of zero.
    """

    def threeSum(self, numbers):
      
        result = []
        combinations = list(itertools.combinations(numbers, 3))
        for i in combinations:
            if sum(i) == 0 and sorted(list(i)) not in result:
                i = list(i)
                i.sort()
                result.append(i)
        return sorted(result)

if __name__ == '__main__':
    nums = [-1,0,1,2,-1,-4]
  
    solution = Solution()
    print(solution.threeSum(nums))

 给一个包含 n 个整数的数组 num，写一个函数找到和与给定整数 target 最接近的三元组，返回这三个数的和。 注意事项： 只需要返回三元组之和，无需返回三元组本身。 格式： 输入行依次输入一个整数数组 num 和一个给定的整数 target，最后输出和与 target 最接近的三元组的和。 样例输入 num = [ -1，2，1，-4 ] target = 1 样例输出 2   // -1 + 2 + 1 = 2 

import itertools


class Solution:
  

    def threeSumClosest(self, numbers, target):
        # write your code here


        combinations = list(itertools.combinations(numbers, 3))
        minNumber = sum(combinations[0])
        for i in combinations:
            if abs(sum(i) - target) < abs(minNumber - target):
                minNumber = sum(i)

        return minNumber


if __name__ == '__main__':
    nums =[-1, 2, 1, -4]

    solution = Solution()
    print(solution.threeSumClosest(nums,1))


  点赞