LeetCode-18.4Sum

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]
]

时间复杂度O(n³)

public List<List<Integer>> fourSum(int[] nums, int target) {//set my
        List<List<Integer>> re = new ArrayList<>();
        //Set<List<Integer>> res = new HashSet<>();
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if(0<i&&nums[i]==nums[i-1]){
                continue;
            }
            for (int j = i+1; j < nums.length; j++) {
                if((i+1)<j&&nums[j]==nums[j-1]){
                    continue;
                }
                int k = j+1;
                int l = nums.length-1;
                while(k<l){
                    if(target==(nums[i]+nums[j]+nums[k]+nums[l])){
                        re.add(Arrays.asList(nums[i],nums[j],nums[k],nums[l]));
                        while(k<l&&nums[k]==nums[k+1])k++;
                        while(k<l && nums[l]==nums[l-1])l--;
                        k++;
                        l--;
                    }
                    else if(target>(nums[i]+nums[j]+nums[k]+nums[l])){
                        k++;
                    }
                    else{
                        l--;
                    }
                }
            }
        }
        return re;
    }

 

该方法不是最优解

首先可以通过判断（可理解为剪枝）提高效率

是否存在低于O(n³)的方法？有待更新

 

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