3Sum Closest——LeetCode

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.



    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

题意就是给定一个数组，一个target，在数组中寻找一个三元组之和最接近target的值，这个题是之前3Sum的变形，不同的是这个题没有重复元素，而且设定解唯一。

我的思路是：跟之前的3sum一样，保留左右各一个pointer，遍历的时候，有left和right两个游标，left就从i+1开始，right就从数组最右length-1开始，如果i、left、right三个元素加起来等target，那么就可以加入结果的List中了，如果sum-target>0，那么说明sum比较大，right应该向左移动，反之left向右移动。

Talk is cheap>>

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

        if (num == null || num.length < 3) {

            return 0;

        }

        int min = Integer.MAX_VALUE;

        int res=0;

        Arrays.sort(num);

        for (int i = 0; i < num.length - 2; i++) {

            int left = i + 1;

            int right = num.length - 1;

            while (left < right) {

                int sum = num[i] + num[left] + num[right];

//                System.out.printf(i+" "+sum);

                if (sum == target) {

                    return target;

                }

                if (min>=abs(sum-target)){

                    min = abs(sum-target);

                    res=sum;

                }

                if (sum-target>0){

                    right--;

                }else {

                    left++;

                }

            }

        }

        return res;

    }

    public int abs(int a){

        return Math.abs(a);

    }