[LeetCode]--18. 4Sum

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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: 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]]

借鉴三个数的方法，固定两个数，再用两个指针去搜寻。

public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        int len = nums.length;        Arrays.sort(nums);        if (len < 4)            return res;        for (int i = 0; i < len - 1; i++) {            if (i > 0 && nums[i] == nums[i - 1])                continue;            for (int j = i + 1; j < len; j++) {                if (j > i + 1 && nums[j] == nums[j - 1])                    continue;                int begin = j + 1, end = len - 1;                while (begin < end) {                    int sum = nums[i] + nums[j] + nums[begin] + nums[end];                    if (sum == target) {                        List<Integer> list = new ArrayList<Integer>();                        list.add(nums[i]);                        list.add(nums[j]);                        list.add(nums[begin]);                        list.add(nums[end]);                        res.add(list);                        begin++;                        end--;                        while (begin < end && nums[begin] == nums[begin - 1])                            begin++;                        while (begin < end && nums[end] == nums[end + 1])                            end--;                    } else if (sum > target)                        end--;                    else                        begin++;                }            }        }        return res;    }}

另外一种就是固定一个数，调用取三个数的算法。

public List<List<Integer>> fourSum1(int[] nums, int target) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        Arrays.sort(nums);        for (int i = 0; i < nums.length; i++) {            int check = target - nums[i];            if (i > 0 && nums[i - 1] == nums[i])                continue;            List<List<Integer>> getThreeSums = threeSum(nums, i + 1,                    nums.length - 1, check);            for (List<Integer> subList : getThreeSums) {                subList.add(nums[i]);                result.add(subList);            }        }        return result;    }    public List<List<Integer>> threeSum(int[] nums, int begin, int finish,            int target) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        for (int i = begin; i <= finish; i++) {            if (i > begin && nums[i - 1] == nums[i])                continue;            int start = i + 1;            int end = finish;            int check = target - nums[i];            while (start < end) {                int twoSum = nums[start] + nums[end];                if (twoSum > check)                    end--;                else if (twoSum < check)                    start++;                else {                    List<Integer> subResult = new ArrayList<Integer>();                    subResult.add(nums[i]);                    subResult.add(nums[start++]);                    subResult.add(nums[end--]);                    result.add(subResult);                    while (start < end && nums[start] == nums[start - 1])                        start++;                    while (start < end && nums[end] == nums[end + 1])                        end--;                }            }        }        return result;    }

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