closest to infinity.nums.i from 0 to n - 3:
left to i + 1 and right to n - 1.left is less than right:
currentSum as nums[i] + nums[left] + nums[right].currentSum is equal to target, return currentSum immediately as it is the closest possible sum.currentSum and target is less than the absolute difference between closest and target, update closest to currentSum.currentSum is less than target, increment left.right.closest after the loop ends.