3Sum Closest

实现：

去重在这里已经不重要了，可以忽略，重复的数不影响最终结果。

记得到初始化int res = numbers[0] + numbers[1] + numbers[numbers.length - 1]，而不是0。和two sum closest不一样，前者是求different。这道题的result是会参与比较影响结果的：Math.abs(sum - target) < Math.abs(res - target)

代码如下:

 public class Solution {  
   /*  
    * @param numbers: Give an array numbers of n integer  
    * @param target: An integer  
    * @return: return the sum of the three integers, the sum closest target.  
    */  
   public int threeSumClosest(int[] numbers, int target) {  
     // write your code here  
      // write your code here  
     Arrays.sort(numbers);  
     int res = numbers[0] + numbers[1] + numbers[numbers.length - 1];  
     for(int i = 0; i < numbers.length - 2; i++) {  
         int left = i + 1;  
         int right = numbers.length - 1;  
         while(left < right) {  
           int sum = numbers[left] + numbers[right] + numbers[i];  
           if (sum < target) {  
             left++;  
           } else {  
             right--;  
           }  
           if(Math.abs(sum - target) < Math.abs(res - target)) {  
             res = sum;  
           }  
         }  
     }  
     return res;  
   }  
 }