LeetCode OJ：3Sum Closest（最接近的三数之和）

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).

求三个数的和sum，使sum与target相差最小。这题实际上跟前面坐的2Sum，以及3Sum相差不是很大，大体的想法是先排序，然后外层的for循环遍历从小到大的数，然后从剩下的数中取三数之和，然后把最小的一步步保存下来，最终将最小的返回就行了。代码如下：

 class Solution {
 public:
     int threeSumClosest(vector<int>& nums, int target) {
         int sz = nums.size();
         sort(nums.begin(), nums.end());//当然了，首先还是要排序的
         int diff = INT_MAX;//这个直接取INT_MAX,这样计算出来的diff一定是比这个要小的
         int tmpDiff;
         int ans;
         int beg, end;
         for (int i = ; i < sz - ; ++i){
             beg = i + , end = sz - ;
             while (beg < end){
                 int sum = nums[i] + nums[beg] + nums[end];
                 if ((tmpDiff = abs(sum - target)) < diff){
                     diff = tmpDiff;
                     ans = sum;
                 }
                 if (sum > target){
                     end--;
                 }
                 else if (sum < target){
                     beg++;
                 }
                 else
                     return sum;//相等的话就直接返回了
             }
         }
         return ans;
     }
 };

大体上就是这样

java版本的如下所示，基本上与上面的没有什么区别：

 public class Solution {
     public int threeSumClosest(int[] nums, int target) {
          int len = nums.length;
          int ret = 0;
          int diff = Integer.MAX_VALUE;
          Arrays.sort(nums);
         for(int i = 0; i < len - 2; ++i){
             int beg = i + 1;
             int end = len - 1;
             while(beg < end){
                 int tmp = nums[i] + nums[beg] + nums[end];
                 int tmpDiff  = Math.abs(tmp - target);
                 if(tmpDiff < diff){
                     ret = tmp;
                     diff = tmpDiff;
                 }
                 if(tmp < target)
                     beg++;
                 else if(tmp > target)
                     end--;
                 else
                     return target;
             }
         }
         return ret;
     }
 }