Description
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.
Example:
Input: nums =
[-2,0,1,3], and target = 2
Output: 2
Explanation: 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?
Solution
Two-pointer, O(n ^ 2), S(1)
虽然nums中可能会有duplicates，但这道题求的是index triplets，所以不需要考虑重复值情况。所以还是挺简单的，用Two-pointer即可解决。注意计算count时需要加k - j而非1，因为所有(j, [j + 1, k])形成的组合都满足条件。
class Solution {
public int threeSumSmaller(int[] nums, int target) {
if (nums == null || nums.length < 3) {
return 0;
}
Arrays.sort(nums);
int count = 0;
int n = nums.length;
for (int i = 0; i < n - 2; ++i) {
int j = i + 1;
int k = n - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k]; // overflow?
if (sum < target) {
count += k - j; // combinations of j and [j + 1, k]
++j;
} else {
--k;
}
}
}
return count;
}
}
