【算法】Three Sum 问题

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:
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)

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class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> results;
sort(nums.begin(), nums.end());
int size = nums.size();
for(int i = 0; i < size - 2; ++i) {
if ( i > 0 && (nums[i] == nums[i-1]) ) continue;
int j = i + 1, k = size - 1;
while( j < k) {
int a = nums[i], b = nums[j], c = nums[k];
int sum = a + b + c;
if (sum == 0) {
vector<int> result {a, b, c};
results.push_back(result);
++j;
--k;
}
else if(sum < 0){
++j;
}
else {
--k;
}
while ( nums[j] == nums[j - 1] && j < k) {
++j;
}
while ( nums[k] == nums[k + 1] && j < k) {
--k;
}
}
}
return results;
}
};
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