【leetcode】Array——4Sum（18）

题目：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)

自己的思路：暴力循环破解，三个指针挨个循环，第四个数用binary search查找。（这样比用4个指针循环要好）

代码：

public List<List<Integer>> fourSum(int[] nums, int target) {Arrays.sort(nums);List<List<Integer>> result = new ArrayList<List<Integer>>();List<Integer> cur = new ArrayList<Integer>();if(nums.length<4)return result;for(int i=0;i<nums.length-3;i++){if(i>0&&nums[i]==nums[i-1])continue;for(int j=i+1;j<nums.length-2;j++){if(j>i+1&&nums[j]==nums[j-1])continue;for(int k=j+1;k<nums.length-1;k++){if(k>j+1&&nums[k]==nums[k-1])continue;int rest = target-nums[i]-nums[j]-nums[k];int restIndex = binarySearch(nums, k+1, rest);if(restIndex!=-1){cur.add(nums[i]);cur.add(nums[j]);cur.add(nums[k]);cur.add(nums[restIndex]);result.add(new ArrayList<Integer>(cur));cur.clear();}}}}System.out.println(result);return result;}private int binarySearch(int[]nums,int start,int rest){int end = nums.length-1;while(start<=end){int mid = start+(end-start)/2;if(nums[mid]==rest)return mid;else if(nums[mid]>rest)end = mid -1;else start=mid+1;}return -1;}

代码虽然被leetcode Accepte了，但是耗时太长。看看下面大神的解法吧！

大神的思路：先针对一个数nums[i]判断该数是否可能，如果nums[i]+max*3<target 或则nums[i]*4>target 等其他异常情况，则直接跳过，变成3Sum问题；同样可以化解为2Sum问题。 算法的时间复杂度O(n^3).

链接：https://leetcode.com/discuss/69517/7ms-java-code-win-over-100%25

代码：

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;    }

另一个思路：用hashtable

map中key为2数之和，value为两数的坐标对的列表。

用两层循环遍历前两个数，后两个数从map中查询

代码：略