Same algorithm as 3sum problem, where we sort `nums`

, then use two pointers to check all the possible combinations, while fixing one element.

In this problem, we just need to add a new variable `diff`

to track the difference between target and current best result. In addition, we move the pointers in terms of `diff`

(be careful with the sign)

```
def threeSumClosest(self, nums, target):
result, diff = 0, sys.maxint
nums.sort()
for i in xrange(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left, right = i + 1, len(nums) - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
hold_diff = abs (total - target)
if not hold_diff:
return total
if hold_diff < diff:
result = total
diff = hold_diff
if total < target:
left += 1
else:
right -= 1
return result
```