leetcode018 Four Sum

最新推荐文章于 2020-08-06 22:36:31 发布

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本文解析了给定整数数组寻找四个数相加等于特定目标值的所有唯一组合的算法。通过对3Sum问题的改进，文章详细介绍了如何通过排序和双指针技巧避免重复解并确保元素按非递减顺序排列。

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题目

18. 4 Sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
- Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
- The solution set must not contain duplicate quadruplets.

For example, given array S = {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)

思路：

根据3Sum进行改进。

代码：

public List<List<Integer>> fourSum(int[] nums, int target)
{
    Arrays.sort(nums);
    List<List<Integer>> lists = new ArrayList<>();
    for(int i = 0; i < nums.length - 3; i++)
    {
        if(i == 0 || nums[i] != nums[i - 1])
        {
            for(int j = i+1; j < nums.length-2; j++)
            {
                if(j == i+1 || nums[j] != nums[j-1])
                {
                    int low = j+1;
                    int high = nums.length-1;
                    int sum = target-nums[i] -nums[j];
                    while(low < high)
                    {
                        if(nums[low]+nums[high] == sum)
                        {
                            lists.add(Arrays.asList(nums[i],nums[j],nums[low],nums[high]));
                            while(low < high && nums[low] == nums[low+1]) low++;
                            while(low < high && nums[high] == nums[high-1]) high--;
                            low++;high--;
                        }
                        else if (nums[low] + nums[high] < sum) low++;
                        else high--;
                    }
                }
            }
        }
    }
    return lists;
}