编程题－三数求和

数组中寻找三数之和为特定目标的组合

最新推荐文章于 2022-10-27 19:14:53 发布

原创最新推荐文章于 2022-10-27 19:14:53 发布 · 1.8k 阅读

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本文介绍了一种算法，用于在给定数组中寻找三个数的组合，其和等于特定目标值。该算法首先构建哈希表统计数组中各元素的出现次数，然后通过遍历不同元素的组合来计算满足条件的组合数量。文章详细解释了算法的实现过程，并提供了一个具体的代码示例。

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# -*- encoding:utf-8 -*-
# A = [1, 1, 2, 2, 2, 2]
# target = 5
from __future__ import print_function

def main(A, target):
    count = 0
    num = len(A)
    hash1 = dict()
    for indx_i in range(num):
        if A[indx_i] in hash1:
            hash1[A[indx_i]] += 1
        else:
            hash1[A[indx_i]] = 1

    #print("hash table: ", hash1)
    keys = list(hash1.keys())

    for indx_i in range(len(keys)):
        i = keys[indx_i]
        if hash1.get(i) > 2 and i + i + i == target:
            count_delta = (hash1.get(i - 2) * hash1.get(i - 1) * hash1.get(i)) / 6
            count += count_delta
            print ("+", i, i, i, ":", count_delta)
        for indx_j in range(indx_i + 1, len(keys)):
            j = keys[indx_j]
            if hash1.get(i) > 1 and i + i + j == target:
                count_delta = (hash1.get(i) * (hash1.get(i) - 1) * hash1.get(j)) / 2
                count += count_delta
                print ("+", i, i, j, ":",  count_delta)

            if hash1.get(j) > 1 and i + j + j == target:
                count_delta = (hash1.get(i) * hash1.get(j) * (hash1.get(j) - 1)) / 2
                count += count_delta
                print ("+", i, j, j, ":",  count_delta)
            for indx_k in range(indx_j + 1, len(keys)):
                k = keys[indx_k]
                if i + j + k == target:
                    count_delta = hash1.get(i) * hash1.get(j) * hash1.get(k)
                    count += count_delta
                    print ("+",i, j, k, ":",  count_delta)
    return count
if __name__ == '__main__':
   result = main(A=[1, 1, 2, 2, 2, 2, 3, 3], target=5)
   print("result:", result)

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