# 923. 3Sum With Multiplicity 三数之和的多种可能

## # 题目描述：

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

## # 解题方法

``````class Solution(object):
def threeSumMulti(self, A, target):
"""
:type A: List[int]
:type target: int
:rtype: int
"""
count = collections.Counter(A)
Aset = set(A)
Alist = list(Aset)
Alist.sort()
_lenA = len(Alist)
res = 0
for i in range(_lenA):
for j in range(i, _lenA):
c = target - Alist[i] - Alist[j]
if c >= Alist[j] and c in Aset:
if Alist[i] != Alist[j] != c:
res += count[Alist[i]] * count[Alist[j]] * count[c]
elif Alist[i] == Alist[j] and Alist[j] != c:
res += count[c] * self.caculate(count[Alist[i]], 2)
elif Alist[j] == c and Alist[i] != Alist[j]:
res += count[Alist[i]] * self.caculate(count[Alist[j]], 2)
elif Alist[i] == c and Alist[j] != c:
res += count[Alist[j]] * self.caculate(count[c], 2)
else:
res += self.caculate(count[Alist[i]], 3)
return res % (10 ** 9 + 7)

def caculate(self, x, i):
if i == 2:
return x * (x - 1) / 2
elif i == 3:
return x * (x - 1) * (x - 2) / 6
``````

## # 日期

2018 年 10 月 14 日 —— 周赛做出来3个题，开心