【leetcode】923. 3Sum With Multiplicity

题目如下：

Given an integer array A, and an integer target, return the number of tuples i, j, k such that i < j < k and A[i] + A[j] + A[k] == target.

As the answer can be very large, return it modulo 10^9 + 7.

Example 1:
Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8

Output: 20

Explanation:

Enumerating by the values (A[i], A[j], A[k]):

(1, 2, 5) occurs 8 times;

(1, 3, 4) occurs 8 times;

(2, 2, 4) occurs 2 times;

(2, 3, 3) occurs 2 times.
Example 2:
Input: A = [1,1,2,2,2,2], target = 5

Output: 12

Explanation:

A[i] = 1, A[j] = A[k] = 2 occurs 12 times:

We choose one 1 from [1,1] in 2 ways,

and two 2s from [2,2,2,2] in 6 ways.
Note:

3 <= A.length <= 3000

0 <= A[i] <= 100

0 <= target <= 300

解题思路：虽然A.length 最大值是300，但是A[i]的值在0~100之间，说明A中有很多重复值，对A去重后length最大值也就100，所以O(n^3)的复杂度完全可以接受。首先对A去重，假设A[i] * A[j] * A[k] == target (i<=j<=k)，那么 A[i] 、A[j]、A[k] 三者之间的值有这么几种情况:

a.三者相等：这种情况，一共存在C(A[i]在A中的个数,3)种组合 (A[i]在A中的个数 >= 3, 这个表达的是A[i]在去重前的A出现的次数)

b.任意两者相等：假设A[i] == A[j] != A[k] ,那么一共存在 C(A[i]在A中的个数,2) * A[k]在A中出现的次数（A[i]在A中的个数,2) >= 2）

c.三者完全不同：这个最简单,一共存在 A[i]在A中出现的次数 * A[j]在A中出现的次数 * A[k]在A中出现的次数

代码如下：

class Solution(object):

    def threeSumMulti(self, A, target):

        """

        :type A: List[int]

        :type target: int

        :rtype: int

        """

        def combination(n,m):

            v1 = 1

            times = 0

            while times < m:

                v1 *= n

                n -= 1

                times += 1

            v2 = 1

            while m > 0:

                v2 *= m

                m -= 1

            return v1 / v2

        dic = {}

        for i in A:

            dic[i] = dic.setdefault(i, 0) + 1

        ul = list(set(A))

        res = 0

        for i in range(len(ul)):

            for j in range(i,len(ul)):

                for k in range(j,len(ul)):

                    if (ul[i] + ul[j] + ul[k]) != target:

                        continue

                    elif ul[i] == ul[j] == ul[k]:

                        if dic[ul[i]] >= 3:

                            res += combination(dic[ul[i]],3)

                    elif ul[i] == ul[j]:

                        if dic[ul[i]] >= 2:

                            res += (combination(dic[ul[i]],2) * dic[ul[k]])

                    elif ul[i] == ul[k]:

                        if dic[ul[i]] >= 2:

                            res += (combination(dic[ul[i]], 2) * dic[ul[j]])

                    elif ul[j] == ul[k]:

                        if dic[ul[j]] >= 2:

                            res += (combination(dic[ul[j]], 2) * dic[ul[i]])

                    else:

                        res +=  (dic[ul[i]] * dic[ul[j]] * dic[ul[k]])

        return res % (pow(10,9) + 7)

巴特西

【leetcode】923. 3Sum With Multiplicity

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