Problem:
Given an array S of n integers, are 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 be 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:
先对数组进行排序，时间复杂度O(log(n))，然后定好一个数的位置，查找另外两个数的和等于-nums[i]的组合，由于数组排好序了，所以可以从两边往中间走，当结果大于0的时候后边往后退一步，否则前边进一步，时间复杂度O(n^2)，所以时间复杂度为O（n^2）
题目大意：
给一组数组，要求得出所有和为0的数字组合，要求数字组合不能重复出现，并且按照升序排列
解题思路：
见Solution.
Java源代码（用时437ms）：
public class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> 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语言源代码（用时48ms）：
/*** Return an array of arrays of size *returnSize.* Note: The returned array must be 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)*3);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++源代码（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源代码（407ms）：
class Solution:# @param {integer[]} nums# @return {integer[][]}def threeSum(self, nums):res = []length=len(nums)if length<3:return resnums.sort()for i in range(length):if nums[i]>0:breakif i>0 and nums[i]==nums[i-1]:continuebegin=i+1;end=length-1while 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-=1while begin<end and nums[begin]==nums[begin-1]:begin+=1while begin<end and nums[end] == nums[end+1]:end-=1elif sum>0:end-=1else:begin+=1return res