leetcode18—4Sum-优快云博客

本文深入探讨了四数之和问题的算法解决方案，通过双指针技巧在已排序数组中寻找四个数相加等于目标值的所有唯一组合，避免了重复结果。该算法基于3Sum的解决思路进行扩展，适用于竞赛编程和面试准备。

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]
]

想法：类似于3Sum的解决方式，设立双指针求解

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int len = nums.size();
        vector<vector<int>> result;
        if( 0 == len)
            return result;
        sort(nums.begin(),nums.end());
        for(int i = 0 ; i < len-3 ; i++){
            if(i != 0 && nums.at(i) == nums.at(i-1))
                continue;
            if( i !=0 && nums.at(i) + nums.at(i+1)+nums.at(i+2)+nums.at(i+3) > target)
                break;
            if( i != 0 && nums.at(i) + nums.at(len-1)+nums.at(len-2)+nums.at(len-3) < target)
                continue;
                for(int j = i+1 ; j < len-2 ; j++){
                    if( j > i+1 && nums[j] == nums[j-1])
                        continue;
                    if(j>i+1 && nums[i] + nums[j] + nums[j+1] + nums[j+2] > target)
                        break;
                    if(j > i+1 && nums[i] + nums[j] + nums[len-1] + nums[len-2] < target)
                        continue;
                        
                    int left = j+1;
                        int right = len-1;
                        while(left < right){
                            int temp = nums.at(i) + nums.at(j) + nums.at(left) + nums.at(right);
                            if(temp == target){
                                result.push_back({nums.at(i),nums.at(j),nums.at(left),nums.at(right)});
                                left++;
                                right--;
                                while(left < right && nums[left] == nums[left-1])//避免重复
                                    left++;
                                while(left < right && nums[right] == nums[right+1])
                                    right--;
                            }else if(temp < target){
                                left++;
                            }else{
                                right--;
                            }
                        }
                    
                        
                }
            
        }
        return result;


    }
};