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)
先对数组排序,然后遍历排序后的数组, 用0减去当前数得到target,然后对后面的数使用双指针,left和right分别指向第一个和最后一个,看这两个指针指向的数相加是否为之前得到的数target,如果相等就找到一组添加到结果中,如果小于left指针右移1位,如果大于,right指针左移1位,然后在相加比较。对于重复的数字可以用while循环跳过。
C++:
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; sort(nums.begin(), nums.end()); for (int k = 0; k < nums.size(); ++k) { if (nums[k] > 0) break; if (k > 0 && nums[k] == nums[k - 1]) continue; int target = 0 - nums[k]; int i = k + 1, j = nums.size() - 1; while (i < j) { if (nums[i] + nums[j] == target) { res.push_back({nums[k], nums[i], nums[j]}); while (i < j && nums[i] == nums[i + 1]) ++i; while (i < j && nums[j] == nums[j - 1]) --j; ++i; --j; } else if (nums[i] + nums[j] < target) ++i; else --j; } } return res; } };