[Leetcode 16, Medium] 3Sum Closest

Problem:

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).

Analysis:

This is an extension of the method in the solution to the problem of 3 sum.

A key point in such solution is to use -1 to indicate a variable, which only takes positive value, is not initialized.
This method is very useful in programming.

Solutions:

C++:

int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        
        int min_sum = 0;
        int min_diff = -1;
        for(int i = 0; i < nums.size() - 2;) {
            int temp_target = target - nums[i];
            int local_min_sum = 0;
            int local_min_diff = -1;
            int start = i + 1;
            int end = nums.size() - 1;
            while(start < end) {
                int temp_sum = nums[start] + nums[end];
                if(local_min_diff == -1 || abs(temp_target - temp_sum) < local_min_diff) {
                    if(temp_target == temp_sum)
                        return nums[i] + temp_sum;

                    local_min_sum = nums[i] + temp_sum;
                    local_min_diff = abs(temp_target - temp_sum);
                }
                
                if(temp_sum < temp_target) {
                    ++start;
                    for(; start < end && nums[start] == nums[start - 1];)
                        ++start;
                } else {
                    --end;
                    for(; end < end && nums[end] == nums[end - 1];)
                        --end;
                }
            }
            
            if(min_diff == -1 || local_min_diff < min_diff) {
                min_sum = local_min_sum;
                min_diff = local_min_diff;
            }
            
            ++i;
            for(; i < nums.size() && nums[i] == nums[i - 1];)
                ++i;
        }
        
        return min_sum;
    }

Java:

Python: