[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;
}
};

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