《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;    }