Leetcode 3Sum (C,c++,java,python)

Problem:

Given an array S of n integers, is there elements a, b, c in S such That a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,C) must is in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2-1-4},    A solution set is:    ( -1, 0, 1)    (-1,-1, 2)

Solution: First sort the array, time complexity O (log (n)), and then set a number of positions, find the other two numbers and equal to the-nums[i] combination, because the arrays are ordered, so you can go from both sides to the middle, when the result is greater than 0 back to step backward, Otherwise the front further, Time complexity O (n^2), so the time complexity of O (n^2)
The main idea: to give a set of arrays, requires that all and 0 of the number combination, requires that the number combination can not be repeated, and in ascending order
Problem-solving ideas: see solution.
Java source code (spents 437ms):

public class Solution {public list<list<integer>> threesum (int[] nums) {List<list<integer&gt        ;> res = new arraylist<list<integer>> ();        int len=nums.length;        if (len<3) return res;        Arrays.sort (Nums);            for (int i=0;i<len;i++) {if (nums[i]>0) break;            if (i>0 && nums[i]==nums[i-1]) continue;            int begin=i+1,end=len-1;                while (begin<end) {int sum=nums[i]+nums[begin]+nums[end];                    if (sum==0) {list<integer> List = new arraylist<integer> ();                    List.add (Nums[i]); List.add (Nums[begin]); List.add (Nums[end]);                    Res.add (list);                    begin++;end--;                    while (Begin<end && nums[begin]==nums[begin-1]) begin++;                while (Begin<end && nums[end]==nums[end+1]) end--;                }else if (sum>0) end--; else 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 first,int end) {int temp,l,r;    if (first>=end) return;    Temp=nums[first];    L=first;r=end;        while (L<r) {while (L<r && nums[r]>=temp) r--;        if (l<r) nums[l]=nums[r];        while (L<r && nums[l]<=temp) l++;    if (l<r) nums[r]=nums[l];    } nums[l]=temp;    QuickSort (NUMS,FIRST,L-1); QuickSort (nums,l+1,end);}    int** threesum (int* nums, int numssize, int* returnsize) {int i,sum,top=-1,begin,end;    int** res= (int**) malloc (sizeof (int*) * (numssize* (numsSize-1) * (numsSize-2))/6);        if (numssize<3) {*returnsize=0;    return res;    } quickSort (nums,0,numssize-1);        for (i=0;i<numssize;i++) {if (nums[i]>0) break;        if (i>0 && nums[i]==nums[i-1]) continue;        begin=i+1;end=numssize-1; while (begin<end) {sum=nums[I]+nums[begin]+nums[end];            if (sum==0) {top++;            res[top]= (int*) malloc (sizeof (int.));            Res[top][0]=nums[i];res[top][1]=nums[begin];res[top][2]=nums[end];            begin++;end--;            while (Begin<end && nums[begin]==nums[begin-1]) begin++;            while (Begin<end && nums[end]==nums[end+1]) end--;            }else if (sum>0) end--;        else begin++;    }} *returnsize=top+1; return res;}

C + + source code (66MS):

Class Solution {public:vector<vector<int>> Threesum (vector<int>& nums) {Vector<vector        <int>> Res;        int len=nums.size ();        if (len<3) {return res;        } sort (Nums.begin (), Nums.end ());            for (int i=0;i<len;i++) {if (nums[i]>0) break;            if (i>0 && nums[i]==nums[i-1]) continue;            int begin=i+1,end=len-1;                while (begin<end) {int sum=nums[i]+nums[begin]+nums[end];                    if (sum==0) {vector<int> T;                    T.push_back (Nums[i]);                    T.push_back (Nums[begin]);                    T.push_back (Nums[end]);                    Res.push_back (t);                    begin++;end--;                    while (Begin<end && nums[begin]==nums[begin-1]) begin++;                while (Begin<end && nums[end]==nums[end+1]) end--; }else if (sum>0) {end--;            }else begin++;    }} return res; }};

Python source code (407MS):

Class solution:    # @param {integer[]} nums    # @return {integer[][]}    def threesum (self, nums):        res = []        Length=len (nums)        if Length<3:return res        nums.sort () for        I in range (length):            if nums[i]>0: Break            if i>0 and Nums[i]==nums[i-1]:continue            begin=i+1;end=length-1 while            begin < end:                sum= Nums[i]+nums[begin]+nums[end]                if sum==0:                    Tmp=[nums[i],nums[begin],nums[end]]                    res.append (TMP)                    begin+=1;end-=1 while                    begin<end and nums[begin]==nums[begin-1]:begin+=1 while                    Begin<end and Nums[end] = = nums[end+1]:end-=1                elif sum>0:end-=1                else:begin+=1        return res

Leetcode 3Sum (C,c++,java,python)