题目：

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1][-2, 0, 3]

Follow up:

Could you solve it in O(n2) runtime?

---

---

 

 

 

 

方法一：

class Solution {public:    int threeSumSmaller(vector
     
      & nums, int target) {        int res = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i < int(nums.size() - 2); ++i) {            int left = i + 1, right = nums.size() - 1, sum = target - nums[i];            for (int j = left; j <= right; ++j) {                for (int k = j + 1; k <= right; ++k) {                    if (nums[j] + nums[k] < sum) ++res;                }            }        }        return res;    }};
     

 

 

 

 

方法二：

• 双指针
  class Solution2 {public:    int threeSumSmaller(vector
         
          & nums, int target) {        if (nums.size() < 3) return 0;        int res = 0, n = nums.size();        sort(nums.begin(), nums.end());        for (int i = 0; i < n - 2; ++i) {            int left = i + 1, right = n - 1;            while (left < right) {                if (nums[i] + nums[left] + nums[right] < target) {                    res += right - left;                    ++left;                } else {                    --right;                }            }        }        return res;    }};