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?
一刷
题解：two pointer
i从0遍历到len-3
left初始化为i+1, right初始化为len-1
如果nums[i] + nums[left] + nums[right] < target， 表示right可以取到所有(left, right]的值，count+= right-left;
否则right--;
public class Solution {
int count;
public int threeSumSmaller(int[] nums, int target) {
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;
}
}
