题目链接在此。上古老题。
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: 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] ]把数组中所有相加为0的三元组找出来，不能有重复。
先排序，然后通过比较相邻元素，利用一切机会去重。2层循环，时间复杂度O(n^2)。
class Solution { public: vector<vector<int> > threeSum(vector<int>& nums) { vector<vector<int> > result; if (nums.size() < 3) return result; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() - 2; i++) { if (i == 0 || nums[i] != nums[i - 1]) { int low = i + 1, high = nums.size() - 1; while (low < high) { if (nums[i] + nums[low] + nums[high] == 0) { int ans[] = { nums[i], nums[low], nums[high] }; result.push_back(vector<int>(ans, ans + 3)); while (low < high && nums[low] == nums[low + 1]) low++; while (low < high && nums[high] == nums[high - 1]) high--; low++; high--; } else if (nums[i] + nums[low] + nums[high] < 0) { while (low < high && nums[low] == nums[low + 1]) low++; low++; } else if (nums[i] + nums[low] + nums[high] > 0) { while (low < high && nums[high] == nums[high - 1]) high--; high--; } } } } return result; } };
