leetcode 18. 4Sum（四个数之和）

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]
]

给出一个数组和一个target，问数组中是否存在4个数字的和为target，找出所有不重复的解

思路：
因为可能有重复解，所以要先排序，跳过重复数字
4 Sum中选出一个数字，剩余部分用3 Sum解决，同理，3 Sum中取一个数字，剩余部分用2 Sum解决
2 Sum用双指针

    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(nums == null || nums.length < 4) {
            return result;
        }
        
        int n = nums.length;
        Arrays.sort(nums);
        
        for(int i = 0; i < n - 3; i ++) {
            if(i > 0 && nums[i] == nums[i-1]) continue;
            if (nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target) break;
            if (nums[nums.length-1] + nums[nums.length-2] + nums[nums.length-3] 
                + nums[nums.length-4] < target) break;
            for(int j = i+1; j < n-2; j++) {
                if(j > i+1 && nums[j] == nums[j-1]) continue;
                int left = j + 1;
                int right = n - 1;
                int sum = target - nums[i] - nums[j];
                while(left < right) {
                    int localSum = nums[left] + nums[right];
                    if(localSum == sum) {
                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        while(left < right && nums[left+1] == nums[left]) {
                            left ++;
                        }
                        while(left < right && nums[right-1] == nums[right]) {
                            right --;
                        }
                        left ++;
                        right --;
                    } else if(localSum < sum) {
                        left ++;
                    } else {
                        right --;
                    }
                }
            }
        }
        return result;
    }