[leetcode]3Sum Closest

至此，终于把LeetCode Old OJ里的132题刷完了，小小的成就。

此题算法简单，就是O(n^2)，采用和Two Sum类似的做法就是了。我的代码略有麻烦之处，主要是在函数里判断了一次abs的差值，外面又判断了一次，但总体不影响。

注意，先选定i，然后在i后面的元素做TwoSum，意思是当第一个元素是i时的结果，这样就不会重复。

也有不需要子函数的写法，更简洁：http://www.cnblogs.com/graph/p/3343847.html

public class Solution {

    public int threeSumClosest(int[] num, int target) {

        Arrays.sort(num);

        int len = num.length;

        int diff = target - (num[0] + num[1] + num[2]);

        for (int i = 0; i < len - 2; i++)

        {

            int tmp = twoSumClosest(num, target - num[i], i + 1);

            if (tmp == 0) return target;

            if (Math.abs(tmp) < Math.abs(diff))

            {

                diff = tmp;

            }

        }

        return target - diff;

    }

    

    private int twoSumClosest(int[] num, int target, int l)

    {

        int r = num.length - 1;

        int diff = target - num[l] - num[r];

        while (l < r)

        {

            int sum = num[l] + num[r];

            if (sum == target)

            {

                return 0;

            }

            else

            {

                int tmp = target - num[l] - num[r];

                if (Math.abs(tmp) < Math.abs(diff))

                {

                    diff = tmp;

                }

                if (sum > target)

                {

                    r--;

                }

                else

                {

                    l++;

                }

            }

        }

        return diff;

    }

}