leecode -- 3sum Closet

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).
class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        sort(num.begin(),num.end());
        int base = 0,left = 1,right = num.size() - 1;
        int minSum = num[base] + num[left] + num[right];int mindistance = abs(minSum - target);
        for(base = 0;base< num.size();base++)
        {
            left = base + 1; right = num.size() -1;
            while(left < right)
            {
                int sum = num[base] + num[left] + num[right];
                int dis = sum - target ;
                if(dis > 0)
                {
                    right--;
                }else if(dis <0)
                {
                    left++;
                }else
                {
                    return sum;
                }
                
                if(abs(dis) < mindistance)
                {
                    minSum = sum;
                    mindistance = abs(dis);
                }
            }
            
        }
        return minSum;
    }
};

 

posted on 2014-05-28 10:32  berkeleysong  阅读(198)  评论(0编辑  收藏  举报

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