摘要：這里需要注意及時處理掉重復(fù)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(shù)組已經(jīng)被排序，所以可以根據(jù)數(shù)組中元素的位置判斷接下來的情況是否有可能合成目標(biāo)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(shù)組中找到所有四個數(shù)字，這四個數(shù)字的和為目標(biāo)值。這四個數(shù)字的組合不能重復(fù)。

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(shù)字，使其和為目標(biāo)值。

在three-sum的基礎(chǔ)上在外圍再加一圈循環(huán)，即在最左側(cè)數(shù)字固定的情況下，尋找右側(cè)數(shù)組中的three-sum結(jié)果。時間復(fù)雜度O(n3)。這里需要注意及時處理掉重復(fù)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(chǔ)上計算4sum不可避免的需要O(n3)的時間復(fù)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(shù)組已經(jīng)被排序，所以可以根據(jù)數(shù)組中元素的位置判斷接下來的情況是否有可能合成目標(biāo)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(fù)
                continue;
            if (z + 3 * max < target) // z is too small 當(dāng)前值即時加上最大值也不足以構(gòu)成目標(biāo)值，進入下一輪循環(huán)
                continue;
            if (4 * z > target) // z is too large 當(dāng)前值*4都大于目標(biāo)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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