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).
题解：
经过几个题的训练，基本上确定是两指针查找问题。找最接近target的三个数之和，第一个想法就是设两个差，一个为正，一个为负，在搜索过程中不断更新，最后比较两个差绝对值的大小，取绝对值小的差，与target相加即可。这里可以在循环内部跳过重复项，也可以不跳过（这样就会进行多余的若干次循环）。
Solution 1 (9ms)
1 class Solution { 2 public: 3 int threeSumClosest(vector<int>& nums, int target) { 4 int diffplus = INT_MAX, diffminus = INT_MIN+1; 5 sort(nums.begin(), nums.end()); 6 int n = nums.size(); 7 for(int i=0; i<n-2; i++) { 8 int j = i + 1, k = n - 1; 9 int a = nums[i]; 10 while(j<k) { 11 int b = nums[j], c = nums[k]; 12 if(a+b+c == target) 13 return target; 14 else if(a+b+c > target) { 15 diffplus = min(diffplus, a + b + c - target); 16 k--; 17 } 18 else { 19 diffminus = max(diffminus, a + b + c - target); 20 j++; 21 } 22 } 23 } 24 return abs(diffminus) < diffplus? target + diffminus : target + diffplus; 25 } 26 };
Solution 1 中我们用了两个变量存储差，那么能不能用一个呢，那么这个diff只能存储差的绝对值，我们怎么知道target该加还是减这个diff呢？解决办法就是在更新diff时同时更新result，在循环内时result == a+b+c;这样就无需target与diff的加减操作了，此时diff的作用只有一个：result是否更新的条件。
Solution 2 (9ms)
1 class Solution { 2 public: 3 int threeSumClosest(vector<int>& nums, int target) { 4 int result = nums[0] + nums[1] + nums[2]; 5 int diff = abs(result - target); 6 sort(nums.begin(), nums.end()); 7 int n = nums.size(); 8 for(int i=0; i<n-2; i++) { 9 int j = i + 1, k = n - 1; 10 while(j<k) { 11 int sum = nums[i] + nums[j] + nums[k]; 12 int now_diff = abs(target - sum); 13 if(now_diff == 0) return target; 14 if(now_diff < diff) { 15 diff = now_diff; 16 result = sum; 17 } 18 else if(sum > target) k--; 19 else j++; 20 } 21 } 22 return result; 23 } 24 };