259. 3Sum Smaller

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;
    }
}