leetcode923 -- 3Sum With Multiplicity

leetcode923 -- 3Sum With Multiplicity

题目

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.

这次是记重数的了,之前第一次的时候并没有记熏数,这次的作法不能够用以前的办法了, 这次可以直接分析这三个数的关系,不需要排序,如果三个数都相等的话,其中两个数相同的话,三个数都不相同的话 代码如下

class Solution(object):
    def threeSumMulti(self, A, target):
        """
        :type A: List[int]
        :type target: int
        :rtype: int
        """
        res = 0
        dic = {}
        for x in A:
            if x not in dic:
                dic[x] = 1
            else:
                dic[x] += 1
        
        for i, x in dic.items():
            for j, y in dic.items():
                k = target -i-j
                if k not in dic:
                    continue
                if i==j==k:
                    res += x*(x-1)*(x-2)/6
                elif i==j and j!= k:
                    res += x*(x-1)/2 *dic[k]
                elif i<j and j<k:
                    res += x*y*dic[k]
                    
        return res%(10**9+7)

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