摘要：這里需要注意及時處理掉重復的情況。那么就需要盡可能排除不可能的情況來提高計算效率。因為數組已經被排序，所以可以根據數組中元素的位置判斷接下來的情況是否有可能合成目標值。

此處為原題地址

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]
]

也就是從數組中找到所有四個數字，這四個數字的和為目標值。這四個數字的組合不能重復。

這題的核心思路請參考我的另一篇博客three-sum，即如何找到三個數字，使其和為目標值。

在three-sum的基礎上在外圍再加一圈循環，即在最左側數字固定的情況下，尋找右側數組中的three-sum結果。時間復雜度O(n3)。這里需要注意及時處理掉重復的情況。

public List> fourSum(int[] nums, int target) {
        List> result = new ArrayList>();
        int length = nums.length;
        if(length<4){
            return result;
        }
        Arrays.sort(nums);
        for(int i = 0 ; i < length-3 ; ){
            int firstValue = nums[i];
            
            //three sum
            for(int j = i+1 ; j < length-2 ; ){
                int secondValue = nums[j];
                int leftPointer = j+1;
                int rightPointer = length-1;
                while(leftPointer=target){
                        while(nums[rightPointer--]==nums[rightPointer] && leftPointer
提高算法的效率
在three-sum的基礎上計算4sum不可避免的需要O(n3)的時間復雜度。那么就需要盡可能排除不可能的情況來提高計算效率。因為數組已經被排序，所以可以根據數組中元素的位置判斷接下來的情況是否有可能合成目標值。
    //思路二：不斷的排除不可能的情況，以加快遍歷，核心的還是3sum
    public List> fourSum2(int[] nums, int target) {
        ArrayList> res = new ArrayList>();
        int len = nums.length;
        if (nums == null || len < 4)
            return res;

        Arrays.sort(nums);

        int max = nums[len - 1];
        //
        if (4 * nums[0] > target || 4 * max < target)
            return res;

        int i, z;
        for (i = 0; i < len; i++) {
            z = nums[i];
            if (i > 0 && z == nums[i - 1])// avoid duplicate 防止重復
                continue;
            if (z + 3 * max < target) // z is too small 當前值即時加上最大值也不足以構成目標值，進入下一輪循環
                continue;
            if (4 * z > target) // z is too large 當前值*4都大于目標值，余下的值不可能再生成最大值
                break;
            if (4 * z == target) { // z is the boundary
                if (i + 3 < len && nums[i + 3] == z)
                    res.add(Arrays.asList(z, z, z, z));
                break;
            }

            threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
        }

        return res;
    }

    /*
     * Find all possible distinguished three numbers adding up to the target
     * in sorted array nums[] between indices low and high. If there are,
     * add all of them into the ArrayList fourSumList, using
     * fourSumList.add(Arrays.asList(z1, the three numbers))
     */
    public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList> fourSumList,
            int z1) {
        if (low + 1 >= high)
            return;

        int max = nums[high];
        if (3 * nums[low] > target || 3 * max < target)
            return;

        int i, z;
        for (i = low; i < high - 1; i++) {
            z = nums[i];
            if (i > low && z == nums[i - 1]) // avoid duplicate
                continue;
            if (z + 2 * max < target) // z is too small
                continue;

            if (3 * z > target) // z is too large
                break;

            if (3 * z == target) { // z is the boundary
                if (i + 1 < high && nums[i + 2] == z)
                    fourSumList.add(Arrays.asList(z1, z, z, z));
                break;
            }

            twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
        }

    }

    /*
     * Find all possible distinguished two numbers adding up to the target
     * in sorted array nums[] between indices low and high. If there are,
     * add all of them into the ArrayList fourSumList, using
     * fourSumList.add(Arrays.asList(z1, z2, the two numbers))
     */
    public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList> fourSumList,
            int z1, int z2) {

        if (low >= high)
            return;

        if (2 * nums[low] > target || 2 * nums[high] < target)
            return;

        int i = low, j = high, sum, x;
        while (i < j) {
            sum = nums[i] + nums[j];
            if (sum == target) {
                fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));

                x = nums[i];
                while (++i < j && x == nums[i]) // avoid duplicate
                    ;
                x = nums[j];
                while (i < --j && x == nums[j]) // avoid duplicate
                    ;
            }
            if (sum < target)
                i++;
            if (sum > target)
                j--;
        }
        return;
    }

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