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?
此题和three sum不同在于要求小于target的所有个数，此题只能用two pointer来做，还有需要注意，此题和three sum不同，个数里面可以包括重复的元素，所以，不需要排除掉相同的数组值组成的和。举例说明：[-2,-2,0,1,3]答案就是有4个了，而不是两个。代码如下：
public class Solution {
public int threeSumSmaller(int[] nums, int target) {
int count = 0;
Arrays.sort(nums);
for(int i=0;i<nums.length-2;i++){
//if(i!=0&&nums[i]==nums[i-1]) continue;
int low = i+1;
int high = nums.length-1;
while(low<high){
if(nums[i]+nums[low]+nums[high]>=target){
high--;
}else{
count+=high-low;
low++;
}
}
}
return count;
}
}