【Leetcode】3Sum Closest

题目链接：https://leetcode.com/problems/3sum-closest/

题目：

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).

思路：

参考http://blog.csdn.net/yeqiuzs/article/details/50272757 解题框架，本题要求跟target距离最近的3元素之和，所以要保存最小距离、最小距离时候的3元素的和。    这里要搞清楚一点，在判断一个已排序数组中是否存在两元素之和 target的时候，要注意可以从两边向中间逼近target，时空复杂度为O(N)，而不是两层for循环两个指针指向两个元素判断是否为target，时空复杂度为O(N^2)。如果是无序数组快排后再向中间逼近是O(N logN)，也比O(N^2）好。  本题解法的时间复杂度为O(N^2)。

算法：

 public int threeSumClosest(int[] nums, int target) {
     Arrays.sort(nums);
     int min = Integer.MAX_VALUE,x = target; //min为target和3个元素和的最小距离，x是3元素和
     for (int i1 = 0; i1 < nums.length; i1++) {
         int i2 = i1 + 1;
         int i3 = nums.length - 1;
         while (i2 < i3) {
             if (Math.abs(nums[i2] + nums[i3] + nums[i1] - target) < min) {
                 min = Math.abs(nums[i2] + nums[i3] + nums[i1] - target);
                 x = nums[i2] + nums[i3] + nums[i1];//保存距离最近的和
             }
             if (nums[i2] + nums[i3] + nums[i1] - target > 0) {
                 i3--;
             } else if (nums[i2] + nums[i3] + nums[i1] - target <= 0) {
                 i2++;
             }
         }
     }
     return x;
 }