题目

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).

分析

和15. 3Sum类似，甚至更加简单。先排序，再固定一个数，对剩下的数进行夹逼操作即可，时间复杂度O(n^2)。

实现

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int ans=nums[0]+nums[1]+nums[2];
        sort(nums.begin(), nums.end());
        for(int i=0; i<nums.size()-2; i++){
            int begin=i+1, end=nums.size()-1;
            while(begin<end){
                int sum = nums[i]+nums[begin]+nums[end];
                if(abs(sum-target)<abs(ans-target))
                    ans = sum;
                if(nums[i]+nums[begin]+nums[end]<target)
                    begin++;
                else
                    end--;
            }
        }
        return ans;
    }
};

思考

不知道为什么我的排名这么靠后，算法都差不多啊。我觉得一定是服务器的问题=_=