[LeetCode] 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).

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 Array Two Pointers

 

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　　这是一道讨论的题目，方法有很多种，最直接的方法是 3层for 循环遍历。

　　第二种方法是 外两次for 循环，第三个数通过 二分查找，时间 是 O（nn logn）

      第三种方法，时间是O(nn)，外层是for循环，内层是 数组首尾指针向中间靠拢，搜索最优结果。

 

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        if(num.size()<4){
            int retNum=0;
            for(int i=0;i<num.size();i++)
                retNum += num[i];
            return retNum;
        }
        vector<int> tmpnum = num;
        sort(tmpnum.begin(),tmpnum.end());
//        for(int i=0;i<tmpnum.size();i++)
//            cout<<tmpnum[i]<<" ";
//        cout<<endl;
        int retNum = tmpnum[0]+tmpnum[1]+tmpnum[2];
        for(int outIdx = 0;outIdx<tmpnum.size();outIdx++){
            int intarget = target-tmpnum[outIdx];
            for(int inleft=0,inright=num.size()-1;inleft<inright;){
                if(inleft==outIdx){
                    inleft++;   continue;
                }
                if(inright==outIdx){
                    inright--;  continue;
                }
//                cout<<tmpnum[inleft]<<" "<<tmpnum[inright]<<endl;
                if(abs(tmpnum[inleft]+tmpnum[inright] + tmpnum[outIdx]-target)<abs(retNum-target) )
                    retNum = tmpnum[inleft]+tmpnum[inright] + tmpnum[outIdx];

                if(tmpnum[inleft]+tmpnum[inright]<intarget)
                    inleft++;
                else if(tmpnum[inleft]+tmpnum[inright]>intarget)
                    inright--;
                else
                    return retNum;
            }
        }
        return retNum;
    }
};

int main()
{
    vector<int > num{1,1,-1,-1,3};
    Solution sol;
    cout<<sol.threeSumClosest(num,-1)<<endl;
    return 0;
}