3sum

Q:

Given an array Sof nintegers, are there elements a, b, cin Ssuch 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)

排序之后，遍历数组每一个元素作为第一个元素，剩下两个元素分别去头和尾向中间滑动。

注意如果一个元素已经被选定为第一个元素，和它相同的值以后都不要再选定为第一个元素了。

A:

class Solution {
 public:
  vector<vector<int> > threeSum(vector<int> &num) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    vector<vector<int> > res;
    if (num.size() < 3) {
      return res;
    }

    sort(num.begin(), num.end());
    res = threeSumInternal(num, 0);
    return res;
  }
 private:
  vector<vector<int> > threeSumInternal(const vector<int> &num, int sum) {
    vector<vector<int> > res;
    int len = num.size();
    for (int i = 0; i < len - 2; ++i) {
      while (i > 0 && i < len && num[i] == num[i - 1]) {
        ++i;
      }
      if (i > len - 2) {
        break;
      }

      int begin = i + 1;
      int end = len - 1;
      while (begin < end) {
        if (num[i] + num[begin] + num[end] == sum) {
          vector<int> part_res;
          part_res.push_back(num[i]);
          part_res.push_back(num[begin]);
          part_res.push_back(num[end]);
          res.push_back(part_res);
          while (begin + 1 < len && num[begin + 1] == num[begin]) {
            ++begin;
          }
          while (end - 1 > 0 && num[end] == num[end - 1]) {
            --end;
          }
          ++begin;
          --end;
        } else if (num[i] + num[begin] + num[end] < sum) {
          ++begin;
        } else {
          --end;
        }
      }
    }
    return res;
  }
};