leetcode: 3Sum Closest

http://oj.leetcode.com/problems/3sum-closest/

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).

思路：

先排序，然后对任意a[i]取j = i + 1，k = S.size() - 1，比较abs(a[i] + a[j] + a[k] - target)，当差小于等于0时，j往后走，反之k往前走。

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int closet_sum = 0, diff = INT_MAX, size = num.size();
        
        sort(num.begin(), num.end());
        
        for (int i = 0; i < size; ++i) {
            int j = i + 1, k = size - 1;
            
            while (j < k) {
                int sum = num[i] + num[j] + num[k];
                int d = abs(sum - target);
                
                if (d < diff) {
                    diff = d;
                    closet_sum = sum;
                }
                
                if (d > 0) {
                    int tmp;
                    
                    if ((sum - target) < 0) {
                        tmp = j + 1;
                        
                        while ((tmp < k) && (num[j] == num[tmp])) {
                            ++tmp;
                        }
                        
                        j = tmp;
                    }
                    else {
                        tmp = k - 1;
                        
                        while ((tmp > j) && (num[tmp] == num[k])) {
                            --tmp;
                        }
                        
                        k = tmp;
                    }
                }
                else {
                    return closet_sum;
                }
            }
        }
        
        return closet_sum;
    }
};