LeetCode 15 3Sum (C,C++,Java,Python)

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 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