4Sum

四数之和算法解析

最新推荐文章于 2019-03-25 12:57:00 发布

转载最新推荐文章于 2019-03-25 12:57:00 发布 · 702 阅读

本文介绍了一种解决四数之和问题的有效算法。该算法通过先排序再使用双指针技巧来查找数组中是否存在四个元素相加等于目标值的情况，并确保结果不包含重复的四元组。适用于寻找所有唯一四元组的场景。

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]
]

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int> > res;
        vector<int> tempres;
        if(nums.empty())
            return res;
        if(nums.size()<=3)
            return res;
        sort(nums.begin(), nums.end());
        for(int j=0; j<nums.size()-3; j++)
        {
            if (j > 0 && nums[j] == nums[j - 1]) continue;   //这一步很关键
            for(int i=j+1; i<nums.size()-2; i++)
            {
                if(i>j+1 && nums[i] == nums[i - 1]) continue;  //这一步也很关键
                int start =i+1;
                int end = nums.size()-1;
                int sum = nums[i];
                while(start<end)
                {
                    if((nums[start]+nums[end]+nums[i]+nums[j])==target)
                    {
                        tempres.push_back(nums[j]);
                        tempres.push_back(nums[i]);
                        tempres.push_back(nums[start]);
                        tempres.push_back(nums[end]);
                        res.push_back(tempres);
                        while( start< end && nums[start] == nums[start + 1]) ++start;
                        while (start< end && nums[end] == nums[end - 1]) --end;
                        tempres.clear();
                        ++start;
                        --end;
                    
                    }else if((nums[start]+nums[end]+nums[i]+nums[j])>target)
                    {
                        --end;
                    }else
                    {
                        ++start;
                    }
                }
            }
                
        }
        return res;
    }
};