[LeetCode] 16. 3Sum Closest 最近三数之和

题目描述

给出含有n个整数的数组s，找出s中和加起来的和最接近给定的目标值的三个整数。返回这三个整数的和。你可以假设每个输入都只有唯一解。

   例如，给定的整数 S = {-1 2 1 -4}, 目标值 = 1.↵↵   最接近目标值的和为 2. (-1 + 2 + 1 = 2).

 

 

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).

这道题让我们求最接近给定值的三数之和，是在之前那道 3Sum 的基础上又增加了些许难度，那么这道题让返回这个最接近于给定值的值，
即要保证当前三数和跟给定值之间的差的绝对值最小，所以需要定义一个变量 diff 用来记录差的绝对值，然后还是要先将数组排个序，
然后开始遍历数组，思路跟那道三数之和很相似，都是先确定一个数，然后用两个指针 left 和 right 来滑动寻找另外两个数，
每确定两个数，求出此三数之和，然后算和给定值的差的绝对值存在 newDiff 中，然后和 diff 比较并更新 diff 和结果 closest 即可，代码如下：

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int res;
        int d = INT_MAX;
        sort(num.begin(),num.end());
        int n = num.size();
        for(int k=0;k<n;++k)
        {
            int t = target-num[k];
            int i=k+1,j=n-1;
            while(i<j)
            {
                int m = t-num[i]-num[j];
                if(abs(m)<abs(d)) d=m;
                if(m>0) i++;
                else j--;
            }
        }
        res=target-d;
        return res;
    }
};

我们还可以稍稍进行一下优化，每次判断一下，当 nums[i]*3 > target 的时候，就可以直接比较 closest 和 nums[i] + nums[i+1] + nums[i+2] 的值，返回较小的那个，因为数组已经排过序了，后面的数字只会越来越大，就不必再往后比较了，参见代码如下：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(closest - target);
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 2; ++i) {
            if (nums[i] * 3 > target) return min(closest, nums[i] + nums[i + 1] + nums[i + 2]);
            int left = i + 1, right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                int newDiff = abs(sum - target);
                if (diff > newDiff) {
                    diff = newDiff;
                    closest = sum;
                }
                if (sum < target) ++left;
                else --right;
            }
        }
        return closest;
    }
};