[LeetCode]3Sum
题目描述:(链接)
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)
解题思路:
先对数组进行排序,然后遍历数组,左右夹逼。
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()); auto last = nums.end(); for (auto i = nums.begin(); i != last - 2; ++i) { if (i > nums.begin() && *i == *(i - 1)) { continue; } auto j = i + 1; auto k = last - 1; while (j < k) { int sum = *i + *j + *k; if (sum < 0) { ++j; while (*j == *(j - 1) && j < k) { ++j; } } else if (sum > 0) { --k; while (*k == *(k + 1) && j < k) { --k; } } else { result.push_back({*i, *j, *k}); ++j; --k; while (*j == *(j - 1) && *k == *(k + 1) && j < k) { ++j; } } } } return result; } };