16. 3Sum Closest -Medium

descrition

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).
 

解析

与 3Sum 的思路一样。不同在于，我们现在希望找到距离 target 最近的数，参看代码。

code

#include <iostream>
#include <vector>
#include <algorithm>
#include <limits>

using namespace std;

class Solution{
public:
    int threeSumClosest(vector<int>& nums, int target){
        sort(nums.begin(), nums.end()); // ascending

        int min_gab = numeric_limits<int>::max();
        int ans = target;

        for(int i=0; i<nums.size(); i++){
            int target_local = target - nums[i];
            int ileft = i + 1;
            int iright = nums.size() - 1;
            while(ileft < iright){ // two pointer searching
                int sum = nums[ileft] + nums[iright];
                if(sum == target_local) // right answer
                    return target;
                if(sum < target_local) // move ileft to increase sum
                    ileft++;
                else // sum > target_local
                    iright--;

                int gab = abs(sum - target_local);
                if(gab < min_gab){
                    ans = sum + nums[i];
                    min_gab = gab;
                }
            }
        }

        return ans;

    }
};

int main()
{
    return 0;
}

作者：fanling999 链接：https://www.cnblogs.com/fanling999/