# 3Sum Smaller

### 描述​

Given an array of n integers nums and an integer 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.

Follow up: Could you solve it in O($n^2$) runtime?

Example 1:

Input: nums = [-2,0,1,3], target = 2

Output: 2

Explanation: Because there are two triplets which sums are less than 2:

[-2,0,1]

[-2,0,3]

Example 2:

Input: nums = [], target = 0

Output: 0

Example 3:

Input: nums = [0], target = 0

Output: 0

Constraints:

• n == nums.length
• 0 <= n <= 300
• -100 <= nums[i] <= 100
• -100 <= target <= 100

### 代码​

# 3Sum Smaller# 先排序，然后双指针左右夹逼# Time Complexity: O(n^2)# Space Complexity: from O(logn) to O(n), depending on the# implementation of the sorting algorithmclass Solution:    def threeSumSmaller(self, nums: List[int], target: int) -> int:        nums.sort()        count = 0        for i in range(len(nums)-2):            count += self.twoSumSmaller(nums, i, target - nums[i])        return count    def twoSumSmaller(self, nums: List[int], i: int, target: int) -> int:        count = 0        left, right = i + 1, len(nums) - 1        while left < right:            if nums[left] + nums[right] < target:                count += right - left                left += 1            else:                right -= 1        return count