PROBLEM:
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] ]
SOLVE:
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
if(nums.size()<3) // 跳过重复的
return res;
sort(nums.begin(),nums.end()); // 排序
for(int i=0;i<nums.size()-2;i++){
if(i>0&&nums[i]==nums[i-1]){
continue;
}
int target=0-nums[i];
int head=i+1;
int tail=nums.size()-1;
// 变为求两个数和为target的种类
while(head<tail){
if(target>nums[head]+nums[tail]){
head++;
}
else if(target<nums[head]+nums[tail]){
tail--;
}
else{
vector<int> unit(3);
unit[0]=nums[i];
unit[1]=nums[head];
unit[2]=nums[tail];
res.push_back(unit);
// 跳过重复的
while(head<tail&&nums[head]==unit[1])head++;
while(head<tail&&nums[tail]==unit[2])tail--;
}
}
}
return res;
}
};