259. 3Sum Smaller

题目：

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 conditionnums[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?

链接： http://leetcode.com/problems/3sum-smaller/

题解：

要求O(n²)求3sum smaller。这里我们依然用类似3Sum的方法，但由于只需要求count，而不用求出每个组合，我们可以作到O(n²)。方法还是用2个指针前后夹逼，当i, lo, hi这个组合满足条件时，在[lo, hi]这个闭合区间内的所有组合也应该满足条件，所以我们这里可以直接count += hi - lo， 然后lo++，增大三个值的和来继续尝试，假如不满足条件，则hi--来缩小三个值的和。

Time Complexity - O(n²)， Space Complexity - O(1)

public class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        if(nums == null || nums.length == 0)
            return 0;
        Arrays.sort(nums);
        int count = 0;

        for(int i = 0; i < nums.length - 2; i++) {
            int lo = i + 1, hi = nums.length - 1;
            while(lo < hi) {
                if(nums[i] + nums[lo] + nums[hi] < target) {
                    count += hi - lo;
                    lo++;
                } else {
                    hi--;
                }
            }
        }

        return count;
    }
}

Reference:

https://leetcode.com/discuss/55602/just-another-pointer-direction-which-think-more-intuitive

https://leetcode.com/discuss/63016/accepted-and-simple-java-solution-with-detailed-explanation

https://leetcode.com/discuss/56164/simple-and-easy-understanding-o-n-2-java-solution

https://leetcode.com/discuss/52424/my-solutions-in-java-and-python

https://leetcode.com/discuss/52362/11-lines-o-n-2-python