[LeetCode] 259. 3Sum Smaller tag: Two pointers

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.

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 <= 3500
• -100 <= nums[i] <= 100
• -100 <= target <= 100

Ideas:  T: O(n ^ 2)    S: O(1)

1. sort nums

2. 建一个two sum function, 如果total < target, count += end - start; start += 1;   else end -= 1

3. for loop, 然后以target - nums[i] 作为two sum function 的target, 最后返回总count数

Code:

class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        n = len(nums)
        nums.sort()
        ans = 0
        for i in range(n - 2):
            ans += self.twoSum(nums, i + 1, n - 1, target - nums[i])
        return ans
    
    
    
    def twoSum(self, nums, start, end, target):
        count = 0
        while start < end:
            if (nums[start] + nums[end] < target):
                count += end - start
                start += 1
            else:
                end -= 1
        return count

标签：count,Smaller,end,target,nums,3Sum,pointers,start,int
来源： https://www.cnblogs.com/Johnsonxiong/p/15171316.html