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?
Solution：two pointers
思路： Just basically same with other "sum"
但要求的是的count，所以count += right - left. 双指针和满足<target的情况下，right左减到底也都满足。然后再升加left，继续此过程。
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
public class Solution {
public int threeSumSmaller(int[] nums, int target) {
int count = 0;
Arrays.sort(nums);
int len = nums.length;
for(int i = 0; i < len - 2; i++) {
int left = i+1, right = len - 1;
while(left < right) {
if(nums[i] + nums[left] + nums[right] < target) {
count += right - left;
left++;
} else {
right--;
}
}
}
return count;
}
}
