LeetCode 658. 找到 K 个最接近的元素（二分查找）

示例 1:

输入: [1,2,3,4,5], k=4, x=3
输出: [1,2,3,4]
 

示例 2:

输入: [1,2,3,4,5], k=4, x=-1
输出: [1,2,3,4]
 

说明:

k 的值为正数，且总是小于给定排序数组的长度。
数组不为空，且长度不超过 10^4
数组里的每个元素与 x 的绝对值不超过 10^4

class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
        if(x <= arr[0])
        	return vector<int>(arr.begin(), arr.begin()+k);
        if(x >= arr[arr.size()-1])
        	return vector<int>(arr.end()-k, arr.end());
        int i, left, right, disLeft, disRight;
        vector<int> ans;
        for(i = 0; i < arr.size(); ++i)
        {
        	if(x == arr[i])
        	{
        		ans.push_back(arr[i]);
        		left = i-1;
        		right = i+1;
        		break;
        	}
        	if(x < arr[i])
        	{
        		left = i-1;
        		right = i;
        		break;
        	}
        }//找x左右的位置
        
        while(left >= 0 && right <= arr.size()-1 && ans.size() < k)
        {
        	disLeft = x - arr[left];//左边距离
        	disRight = arr[right] - x;//右边距离
        	if(disLeft <= disRight)
        	{
        		ans.push_back(arr[left]);
        		--left;
        	}
        	else
        	{
        		ans.push_back(arr[right]);
        		++right;
        	}
        }
        if(ans.size() != k)//一边到头了
        {
        	if(left == -1)
        	{
        		while(ans.size() != k)
        			ans.push_back(arr[right++]);
        	}
        	else//right = ans.size()
        	{
        		while(ans.size() != k)
        			ans.push_back(arr[left--]);

        	}
        }
        sort(ans.begin(), ans.end());
        return ans;
    }
};

class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
		int lo = 0, hi = arr.size() - k;
		while (lo < hi) {
			int mid = lo + (hi - lo >> 1);
			if (abs(x - arr[mid]) > abs(arr[mid + k] - x ))
				lo = mid + 1;
			else
				hi = mid;
		}
		return vector<int>(arr.begin() + lo, arr.begin() + lo + k);
	}
};