Leetcode 18.4Sum (4 numbers and) ideas and methods of solving problems

4Sum
Given an array S of n integers, is 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 is 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)

Idea: This problem and 3Sum is related to the problem, the problem is converted into b+c+d = Target-a =k 3Sum.

Specific methods and ideas see the following code:

    Public list<list<integer>> foursum (int[] nums, int target) {list<list<integer>> List =        New Arraylist<list<integer>> ();        int len = nums.length;        if (Len <= 3)//insufficient length, directly return to the return list; Arrays.sort (nums);//Sort//Start loop for (int i = 0; i < len-2; i++) {//if target> 0 and Nums[i]>targe T, the remaining array addition cannot be =target if (Target > 0 && nums[i] > target) | |                (Target < 0 && nums[len-1] < target))                        Break                        if (i > 0 && nums[i] = = Nums[i-1])//eliminate repetitive continue;            Converted from a+b+c+d = T to B + C + d = t-a int a = target-nums[i]; for (int j = i + 1; j < len-1;j++) {if (a > 0 && nums[j] > a) | | (A < 0 && nums[len-1] < a))                            If Nums[j]>a, then the remaining array addition cannot =a break; if (J > i+1 && nums[J] = = Nums[j-1]) continue;                int m = j+1;                                int n = len-1;                    while (M < n) {int k = Nums[j] + nums[m] + nums[n];                        if (k = = a) {list<integer> Al = new Arraylist<integer> ();                        Al.add (Nums[i]);                        Al.add (Nums[j]);                        Al.add (Nums[m]);                        Al.add (Nums[n]);                        List.add (AL);                        m++;                        n--;                        while (M < n && nums[m] = = nums[m-1]) m++;                    while (M < n && nums[n] = = nums[n+1]) n--;                        } else{//Change position mark if (K < a) m++;                    else n--;}}}} return list; }

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Leetcode 18.4Sum (4 numbers and) ideas and methods of solving problems