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[LintCode] 3Sum 三数之和
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.
Notice
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.
Have you met this question in a real interview?
Yes
Example
For example, given array S = {-1 0 1 2 -1 -4}, A solution set is:
(-1, 0, 1)(-1, -1, 2)
LeetCode上的原题，请参见我之前的博客3Sum。
class Solution {public: /** * @param numbers : Give an array numbers of n integer * @return : Find all unique triplets in the array which gives the sum of zero. */ vector<vector<int> > threeSum(vector<int> &nums) { vector<vector<int>> res; sort(nums.begin(), nums.end()); for (int k = 0; k < nums.size() - 2; ++k) { if (nums[k] > 0) break; if (k > 0 && nums[k] == nums[k - 1]) continue; int target = 0 - nums[k], 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; }};
[LintCode] 3Sum 三数之和
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