【LeetCode】 4Sum 【Python C++】

题目：

从给定的一堆数字中找出 所有 4个数之和=taget的组合。组合与组合之间不得重复

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

解答：

#C++版本（观摩了 leetcode讨论区的大佬）
class K_Sum{
public:
    vector<vector<int>> k_Sum(vector<int>& nums,int target,int k){
        if(nums.size()<k){
            return {};
        }
        sort(nums.begin(),nums.end());
        return k_sum_helper(nums,target,k,0);
    
}
private:
    vector<vector<int>> k_sum_helper(vector<int>& nums,int target,int k,int begin_index){
        if(begin_index == nums.size() || k*nums[begin_index] > target || k*nums[nums.size()-1] <target){
            return {};
        }
        vector<vector<int>> result;
        if (k == 2) {
            for (int left = begin_index, right = nums.size() - 1; left < right;
                    ++left) {
                while (left < right && nums[left] + nums[right] > target) {
                    --right;
                }
                if (left == right) return result;
                if (nums[left] + nums[right] == target) {
                    result.push_back({nums[left], nums[right]});
                }
                else if (right != nums.size() - 1) ++right;
                while (left < right && nums[left + 1] == nums[left]) ++left;
            }
            return result;
        }
        for(int i=begin_index;i+k<=nums.size();i++){
            auto temp_result=k_sum_helper(nums,target-nums[i],k-1,i+1);
            for(auto & v :temp_result){
                v.push_back(nums[i]);
                result.emplace_back(std::move(v));
            }
            while(i+k<nums.size() && nums[i]==nums[i+1]) i++;
        }
        return result;
        
    }
};


class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        return K_Sum().k_Sum(nums, target, 4);
    }
};

#Python 版本
class Solution:
    def fourSum(self, nums, target):
        ans=[]
        nums.sort()
        for i in range(0,len(nums)):
            for j in range(i+1,len(nums)):
                k,z=j+1,len(nums)-1
                while k<z:
                    sum_=nums[i]+nums[j]+nums[k]+nums[z]
                    if sum_==target:
                        ans_1=[nums[i],nums[j],nums[k],nums[z]]
                        if ans_1 not in ans:
                            ans.append(ans_1)
                            k+=1
                            z-=1
                            continue
                        else:
                            k+=1
                            z-+1
                    else:
                        if sum_>target:
                            z-=1
                        else:
                            k+=1
    
                
        return ans