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)
两个数之和等于一个数的问题扩展。
先排序，依次遍历每个数a[i]，在后面的数组中找到两个数之和等于-a[i]。
class Solution {
public:
vector<vector<int>> threeSum(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int>> result;
auto n = num.size();
size_t i;
for (i = 0; i < n; ++i) {
if (i > 0 && num[i] == num[i - 1])
continue;
int a = num[i];
int low = i + 1, high = n - 1;
while (low < high) {
int b = num[low];
int c = num[high];
if (a + b + c == 0) {
vector<int> v = {a, b, c};
result.push_back(v);
while (low + 1 < n && num[low] == num[low + 1]) low++;
while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
low++;
high--;
} else if (a + b + c > 0) {
while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
high--;
} else {
while (low + 1 < n && num[low] == num[low + 1]) low++;
low++;
}
}
}
return result;
}
};