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?

Tips:

O(n^2). 2 points. 先固定一个数i, 然后设left和right搜，如果满足条件则说明left 和 [left + 1, right]之间的任意数都满足，left++再看； 如果不满足则right--再看。

Code：

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;
    }
}