Given an array nums
of n integers, are there elements a, b, c in nums
such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
解题思路:先将数组排序,然后开始遍历,先确定一个数,遇到这个数与前面的数相同时,则跳过,然后用双指针法,在该数的右边找两个数使得这三个数相加为0;
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(),nums.end());
vector<vector<int>>result;
for(int i=0;i<nums.size();i++){
while(i!=0&&nums[i-1]==nums[i]) i++;
int sum=0-nums[i];
int low=i+1,high=nums.size()-1;
while(low<high){
int temp=nums[low]+nums[high];
if(temp==sum){
vector<int>tempresult(3);
tempresult[0]=nums[i];
tempresult[1]=nums[low++];
tempresult[2]=nums[high--];
result.push_back(tempresult);
while(low<high&&nums[low]==tempresult[1]) low++;
while(low<high&&nums[high]==tempresult[2]) high--;
}
else if(temp>sum)
high--;
else
low++;
}
}
return result;
}
};