K Closest Numbers In Sorted Array

K Closest Numbers In Sorted Array.png

===================== 解題思路 =====================

二分法先找到最接近的點做為起點（或是跟 target 相同點) 然後以此點的左右鄰點開始比較跟起點的距離 比較小的就存入 然後移動鄰點往兩邊擴散繼續做下一輪的比較

===================== C++ code ====================

<pre><code>
class Solution {

public:

/**
 * @param A an integer array
 * @param target an integer
 * @param k a non-negative integer
 * @return an integer array
 */

int findMin(vector<int> &A, int &left, int &right, int target)
{
    if(left < 0) return right;
    if(right >= A.size()) return left;
    return abs(A[left] - target) <= abs(A[right] - target)? left : right;
}

vector<int> kClosestNumbers(vector<int>& A, int target, int k) {
    // Write your code here
    
    int left = 0, start = -1, right = A.size() -1;
    while(left + 1 < right)
    {
        int mid = (left + right) / 2;
        if(A[mid] == target){
            start = mid;
            break;
        }
        if(A[mid] > target) right = mid;
        else left = mid;
    }
    if(start == -1)
    {
        start = findMin(A, left, right, target);
    }
    
    vector<int> res;
    left = start -1;
    right = start;
    for(int i = 0; i < k; i++)
    {
        int tmp = findMin(A, left, right, target);
        if(tmp == left) left--;
        else right++;
        res.emplace_back(A[tmp]);
    }
    return res;
}

};
<code><pre>