Note:
For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0, 1) (-1, -1, 2)
思路:
先回忆一下2sum问题的解法。对从小到大排序的数组,在数组前后各放一个指针。如果指针对应数的和大于给定值,则右移最左边的指针,反之左移最右边的指针,直到和是给定值或者两个指针碰撞。3Sum可以首先固定一个数,然后进行2Sum。
如果需要unique的解法,一种方法是先给出所有解,然后做一个std::unique,另一种方法是记录之前指针的值,然后进行新的循环时检测指针对应的值是否更新。
题解:
class Solution { public: vector<vector<int>> ret; int head; void two_sum (const vector<int>& v, vector<int>::const_iterator c1, int expectation) { if (v.cend() - c1 <= 1) return; vector<int>::const_iterator c2 = v.cend() - 1; while (c1 < c2) { int twos = *c1 + *c2; if (twos == expectation) { ret.emplace_back (vector<int> {{head, *c1, *c2}}); int cc1 = *c1; while (c1 < c2 && cc1 == *c1) ++c1; } else if (twos < expectation) ++c1; else --c2; } } vector<vector<int> > threeSum (vector<int>& num) { ret.clear(); if (num.empty()) return ret; sort (begin (num), end (num)); head = num.front(); for (auto n = num.cbegin(); n != num.cend(); ++n) { if (n != num.cbegin() && *n == head) continue; head = *n; two_sum (num, n + 1, -*n); } return ret; } };