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描述
Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:
- nums[a] + nums[b] + nums[c] == nums[d], and
- a < b < c < d
Example 1:
Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.
Example 2:
Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].
Example 3:
Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5
Note:
4 <= nums.length <= 50
1 <= nums[i] <= 100
解析
根据题意，给出一个从 0 开始索引的列表 nums ，题目要求我们返回不同四元组 ( a , b , c , d ) 的数量，使得：
- nums[a] + nums[b] + nums[c] == nums[d]
- a < b < c < d
最直接的办法就是使用暴力解法，思路如下：
- 初始化计数器 result 为 0
- 使用内置函数 combinations(nums, 4) 将所有四元组合都找出来，然后遍历所有的组合判断是否符合 cb[0] + cb[1] + cb[2] == cb[3] ，如果符合就计数器 result 加一
- 遍历结束返回 result
解答
class Solution(object):
def countQuadruplets(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = 0
for cb in combinations(nums, 4):
if cb[0] + cb[1] + cb[2] == cb[3]:
result += 1
return result
运行结果
Runtime: 1780 ms, faster than 9.01% of Python online submissions for Count Special Quadruplets.
Memory Usage: 13.3 MB, less than 74.59% of Python online submissions for Count Special Quadruplets.
解析
另外还可以用四层循环，寻找所有四个索引的组合来找出满足题意的四元组合，这也是一种暴力解法，只不过写法不同，运行的结果和上面的差不多。
解答
class Solution(object):
def countQuadruplets(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = 0
for i in range(len(nums)-3):
for j in range(i+1, len(nums)-2):
for k in range(j+1, len(nums)-1):
for l in range(k+1, len(nums)):
if nums[i]+nums[j]+nums[k] == nums[l]:
result += 1
return result
运行结果
Runtime: 1588 ms, faster than 37.70% of Python online submissions for Count Special Quadruplets.
Memory Usage: 13.4 MB, less than 74.59% of Python online submissions for Count Special Quadruplets.
解析
看了有论坛的大神用到了字典的解法很巧妙，主要是使用了 nums[a] + nums[b] == nums[d] - nums[c] 这一等式，将 nums[a] + nums[b] 保存入字典中，只需要遍历 c 和 d 的索引即可，时间复杂度可以降到 O(n^2) ，性能提升可是太大了！！
解答
class Solution(object):
def countQuadruplets(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
res = 0
d = defaultdict(int)
d[nums[0] + nums[1]] = 1
for i in range(2, n - 1):
for j in range(i + 1, n):
res += d[nums[j] - nums[i]]
for j in range(i):
d[nums[i] + nums[j]] += 1
return res
运行结果
Runtime: 48 ms, faster than 99.18% of Python online submissions for Count Special Quadruplets.
Memory Usage: 13.3 MB, less than 93.44% of Python online submissions for Count Special Quadruplets.
原题链接：leetcode.com/problems/co…
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