[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.
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 = ; k < nums.size() - ; ++k) {
if (nums[k] > ) break;
if (k > && nums[k] == nums[k - ]) continue;
int target = - nums[k], i = k + , j = nums.size() - ;
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 + ]) ++i;
while (i < j && nums[j] == nums[j - ]) --j;
++i; --j;
} else if (nums[i] + nums[j] < target) ++i;
else --j;
}
}
return res;
}
};
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