[array] leetCode-16. 3Sum Closest -Medium

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;
}