leetcode 15

三数相加得一数，分析：先排序然后左右夹逼，复杂度O(n2)。

代码：
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());
const int target = 0;
auto last = nums.end();
for (auto i = nums.begin(); i < last-2; ++i) {
auto j = i+1;
if (i > nums.begin() && *i == *(i-1)) continue;
auto k = last-1;
while (j < k) {
if (*i + *j + *k < target) {
++j;
while(*j == *(j - 1) && j < k) ++j;
} else if (*i + *j + *k > target) {
--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;
}
};