動機

有剪枝差太多

Problem

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

    t
  • 0 <= a, b, c, d < n
    • t
  • a, b, c, and d are distinct.
    • t
  • nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

 

Example 1:

Input: nums = [1,0,-1,0,-2,2], target = 0Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

Example 2:

Input: nums = [2,2,2,2,2], target = 8Output: [[2,2,2,2]]

 

Constraints:

    t
  • 1 <= nums.length <= 200
    • t
  • -109 <= nums[i] <= 109
    • t
  • -109 <= target <= 109

Sol

在k是大於2時，就用backtrack去生剩下的(k-1)sum，最後就是2sum

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        def sumk(ns, target, k):
            if len(ns) <= 1 or k <= 1 or ns[0]*k > target or target > ns[-1] *k:
                return []
            elif k == 2:
                return sum2(ns,target)
            else:
                ret = []
                for i in range(len(ns)):
                    if i == 0 or ns[i-1] != ns[i]:
                        subs = sumk(ns[i+1:],target-ns[i], k-1)
                        if subs:
                            ret += [[ns[i]]+sub for sub in subs]
                return ret
        def sum2(ns,target):
            ret = []
            a,b = 0, len(ns)-1
            while a<b:
                cnt = ns[a]+ns[b]
                if cnt == target:
                    ret.append([ns[a],ns[b]])
                    while a <= b-1 and ns[b-1] == ns[b]:
                        b -= 1
                    while a+1 <= b and ns[a+1] == ns[a]:
                        a += 1
                    a,b = a+1, b-1
                elif cnt > target:
                    while a <= b-1 and ns[b-1] == ns[b]:
                        b -= 1
                    b -= 1
                else:
                    while a+1 <= b and ns[a+1] == ns[a]:
                        a += 1
                    a += 1
            return ret
        
        nums.sort()
        return sumk(nums,target,4)