[leetcode] 454. 4Sum II @ python

原题

https://leetcode.com/problems/4sum-ii/
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:

1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

解法1

构建一个字典, 遍历A和B, 记录a+b的值和个数, 然后遍历C和D, 查看c+d的相反数是否在字典里, 如果在就累加计数.
Time: 2*O(n)
Space: O(1)

代码

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        # use hashtable to store the number and count
        h = {}
        for a in A:
            for b in B:
                h[a+b] = h.get(a+b, 0) + 1
        
        count = 0
        for c in C:
            for d in D:
                if -(c+d) in h:
                    count += h[-(c+d)]
        return count

解法2

构造两个字典, 分别储存a+b, c+d的和与对应的计数, 然后遍历字典, 排列组合求解.

代码

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        d1 = collections.defaultdict(int)
        d2 = collections.defaultdict(int)
        for a in A:
            for b in B:
                d1[a+b] += 1
        
        for c in C:
            for d in D:
                d2[c+d] += 1
                
        count = 0
        for k in d1:
            if -k in d2:
                # we found a match
                count += d1[k]*d2[-k]
                
        return count