Leetcode 4Sum (C,c++,java,python)

Problem:

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)

Solution: With 15 3Sum very similar, but more than a heavy cycle, see: Click to open the link
To give an array, and a target integer, the array can make and for all combinations of target a,b,c,d and satisfy A<=b<=c<=d
Java source Code (527MS):

public class Solution {public list<list<integer>> foursum (int[] nums, int target) {list<list&lt        ;integer>> res=new arraylist<list<integer>> ();        int length=nums.length;        if (length<4) return res;        Arrays.sort (Nums);            for (int i=0;i<length-3;i++) {if (i>0 && nums[i]==nums[i-1]) continue;                for (int j=i+1;j<length-2;j++) {if (j>i+1 && nums[j]==nums[j-1]) continue;                int begin=j+1,end=length-1;                    while (begin<end) {int sum=nums[i]+nums[j]+nums[begin]+nums[end];                        if (sum==target) {list<integer> temp = new arraylist<integer> ();                        Temp.add (Nums[i]);                        Temp.add (Nums[j]);                        Temp.add (Nums[begin]);                        Temp.add (Nums[end]);                        Res.add (temp); begin++;end--;                        while (Begin<end && nums[begin]==nums[begin-1]) begin++;                    while (Begin<end && nums[end]==nums[end+1]) end--;                        }else if (sum>target) {end--;                    while (Begin<end && nums[end]==nums[end+1]) end--;                        }else{begin++;                    while (Begin<end && nums[begin]==nums[begin-1]) begin++;    }}}} return res; }}

C Language Source code (spents 48ms):

/** * Return An array of arrays of size *returnsize. * Note:the returned array must is malloced, assume caller calls free ().    */void QuickSort (int* nums,int start,int end) {int l=start,r=end,tmp=nums[l];    if (start>=end) return;        while (L<r) {while (L<r && nums[r]>=tmp) r--;        if (l<r) nums[l]=nums[r];        while (L<r && nums[l]<=tmp) l++;    if (l<r) nums[r]=nums[l];    } nums[l]=tmp;    QuickSort (NUMS,START,L-1);    QuickSort (Nums,l+1,end);     }int** foursum (int* nums, int numssize, int target, int* returnsize) {int** res = (int**) malloc (sizeof (int*) *1000000);    int i,j,sum,begin,end,*temp;    *returnsize=0;    if (numssize<4) return res;    QuickSort (nums,0,numssize-1);        for (i=0;i<numssize-3;i++) {if (i>0 && nums[i]==nums[i-1]) continue;            for (j=i+1;j<numssize-2;j++) {if (j>i+1 && nums[j]==nums[j-1]) continue;            begin=j+1;end=numssize-1; while (begin<End) {Sum=nums[i]+nums[j]+nums[begin]+nums[end];                    if (sum==target) {temp= (int*) malloc (sizeof (int));                    Temp[0]=nums[i];temp[1]=nums[j];temp[2]=nums[begin];temp[3]=nums[end];                    Res[*returnsize]=temp;                    (*returnsize) + +;                    begin++;end--;                    while (Begin<end && nums[begin]==nums[begin-1]) begin++;                while (Begin<end && nums[end]==nums[end+1]) end--;                    }else if (sum>target) {end--;                while (Begin<end && nums[end]==nums[end+1]) end--;                    } else{begin++;                while (Begin<end && nums[begin]==nums[begin-1]) begin++; }}}} return res;}

C + + source code (141MS):

Class Solution {public:vector<vector<int>> Foursum (vector<int>& nums, int target) {Vecto        r<vector<int>> Res;        int length=nums.size ();        if (length<4) return res;        Sort (Nums.begin (), Nums.end ());            for (int i=0;i<length-3;i++) {if (i>0 && nums[i]==nums[i-1]) continue;                for (int j=i+1;j<length-2;j++) {if (j>i+1 && nums[j]==nums[j-1]) continue;                int begin=j+1,end=length-1;                    while (begin<end) {int sum=nums[i]+nums[j]+nums[begin]+nums[end];                        if (sum==target) {vector<int> temp;                        Temp.push_back (Nums[i]);                        Temp.push_back (Nums[j]);                        Temp.push_back (Nums[begin]);                        Temp.push_back (Nums[end]);                        Res.push_back (temp);                        begin++;end--;while (Begin<end && nums[begin]==nums[begin-1]) begin++;                    while (Begin<end && nums[end]==nums[end+1]) end++;                        }else if (sum>target) {end--;                    while (Begin<end && nums[end]==nums[end+1]) end--;                        }else{begin++;                    while (Begin<end && nums[begin]==nums[begin-1]) begin++;    }}}} return res; }};

Python source code (1842MS):

Class Solution: # @param {integer[]} nums # @param {integer} target # @return {integer[][]} def foursum (self, Nums, target): Length=len (nums) res=[] If Length<4:return res nums=sorted (nums) for                I in range (length-3): If I>0 and Nums[i]==nums[i-1]:continue for J in Range (I+1,length-2):  If J>i+1 and nums[j]==nums[j-1]:continue begin=j+1;end=length-1 while begin <                        End:sum=nums[i]+nums[j]+nums[begin]+nums[end] If Sum==target: Temp=[nums[i],nums[j],nums[begin],nums[end]] Res.append (temp) begin+=1;en D-=1 while Begin<end and nums[begin]==nums[begin-1]:begin+=1 while begin&                        Lt;end and Nums[end]==nums[end+1]:end-=1 elif sum>target:end-=1 While BEGIN&LT;end and Nums[end]==nums[end+1]:end-=1 else:begin+=1 WH Ile Begin<end and Nums[begin]==nums[begin-1]:begin+=1 return res

Leetcode 4Sum (C,c++,java,python)