《LeetBook》leetcode题解(18) : 4Sum[M]

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018. 4Sum

问题

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:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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)

思路

好了，2SUM，3SUM，3SUM Closet 都解决了，我们顺利迎来了最后一个boss，4SUM，但是我们已经见怪不怪了。老思路，把4SUM问题变成3SUM问题。
- 先排序
- 确定一个数，然后文件顺利变成3SUM问题，然后就是完完全全3SUM的解决思路了。

for(int i = 0;i < nums.length-3;i++)
{
     int target_i = target - nums[i];
     if(i > 0 && nums[i] == nums[i-1])  //排除一样的数
                continue;
     ……//3SUM问题
}

整体代码

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {

        Arrays.sort(nums);
        List<List<Integer>> list;
        list = new ArrayList<List<Integer>>();
        int mid,right;
        for(int i = 0;i < nums.length-3;i++)
        {
            int target_i = target - nums[i];
            if(i > 0 && nums[i] == nums[i-1])
                continue;
            for (int left = i+1; left < nums.length-2; left++)
            {
                mid = left+1;
                right = nums.length-1;
                int tmp = target_i-nums[left];
                if(left > i+1 && nums[left] == nums[left-1])
                    continue;
                while(mid < right)
                {
                    if(nums[mid] + nums[right] == tmp)
                    {
                        int tmp_mid = nums[mid],tmp_right= nums[right];
                        list.add(Arrays.asList(nums[i],nums[left], nums[mid], nums[right]));
                        while(mid < right && nums[++mid] == tmp_mid);
                        while(mid < right && nums[--right] == tmp_right);
                    }
                    else if(nums[mid] + nums[right] < tmp)
                        mid++;
                    else
                        right--;
                }
             }
        }
        return list;
    }
}

思路2：排除不可能情况

讨论区rikimberley有个高分的答案，具体思路还是和上面的思路一样的，但是它的运行速度超过了100%的人，是因为它在运行的时候，排除了很多不可能的情况。假设我们考虑4个数分别为A B C D（有序），最大值MAX。
1. A太大，退出：（如果4*A > target）
2. A太小，跳过：（A+4*MAx < target）
3. 确定A后求BCD的3SUM问题
4. B太大，退出：（如果3*B > target）
5. B太小，跳过：（B+3*MAx < target）
……

public List<List<Integer>> fourSum(int[] nums, int target) {
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        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
                continue;
            if (z + 3 * max < target) // z is too small
                continue;
            if (4 * z > target) // z is too large
                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<List<Integer>> 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<List<Integer>> 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;
    }