linear-list/array/3sum-smaller
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2023-12-01
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
分析
先排序，然后双指针左右夹逼，复杂度 $$O(n^2)$$。
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 algorithm
class 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
// 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 algorithm
class Solution {
public int threeSumSmaller(int[] nums, int target) {
if(nums.length < 3) return 0;
Arrays.sort(nums);
int count = 0;
for (int i = 0; i < nums.length - 2; i++) {
count += twoSumSmaller(nums, i, target - nums[i]);
}
return count;
}
private int twoSumSmaller(int[] nums, int i, int target) {
int count = 0;
int left = i + 1, right = nums.length - 1;
while (left < right) {
if (nums[left] + nums[right] < target) {
count += right - left;
left++;
} else {
right--;
}
}
return count;
}
}
// 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 algorithm
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
if(nums.size() < 3) return 0;
sort(nums.begin(), nums.end());
int count = 0;
for (int i = 0; i < nums.size() - 2; i++) {
count += twoSumSmaller(nums, i, target - nums[i]);
}
return count;
}
private:
int twoSumSmaller(const vector<int>& nums, int i, int target) {
int count = 0;
int left = i + 1, right = nums.size() - 1;
while (left < right) {
if (nums[left] + nums[right] < target) {
count += right - left;
left++;
} else {
right--;
}
}
return count;
}
};