题目描述
给定一个排序好的数组,两个整数 k 和 x,从数组中找到最靠近 x(两数之差最小)的 k 个数。返回的结果必须要是按升序排好的。如果有两个数与 x 的差值一样,优先选择数值较小的那个数。
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]
提示
- 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
题解
滑动窗口
代码
滑动窗口(c++)
class Solution {
public:
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int n = arr.size();
int l = 0, r = n-1;
while (r-l >= k) {
if (x-arr[l] <= arr[r]-x) r--;
else l++;
}
vector<int> res(k);
copy(arr.begin()+l, arr.begin()+l+k, res.begin());
return res;
}
};
二分+滑动窗口(c++)
class Solution {
public:
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int n = arr.size();
int l = 0, r = n-1;
while (l < r) {
int m = (l + r) / 2;
if (arr[m] < x) l = m + 1;
else r = m;
}
r = min(n-1, l+k-1);
l = max(0, l-k);
while (r-l >= k) {
if (x-arr[l] <= arr[r]-x) r--;
else l++;
}
vector<int> res(k);
copy(arr.begin()+l, arr.begin()+l+k, res.begin());
return res;
}
};
二分(c++)
class Solution {
public:
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int n = arr.size();
int l = 0, r = n-k;
while (l < r) {
int m = (l + r) / 2;
if (x-arr[m] > arr[m+k]-x) l = m + 1;
else r = m;
}
vector<int> res(k);
copy(arr.begin()+l, arr.begin()+l+k, res.begin());
return res;
}
};
滑动窗口(python)
class Solution:
def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
n = len(arr)
l, r = 0, n-1
while r-l >= k:
if x-arr[l] <= arr[r]-x:
r -= 1
else:
l += 1
return arr[l:l+k]
二分+滑动窗口(python)
class Solution:
def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
n = len(arr)
l, r = 0, n-1
while l < r:
m = (l + r) // 2
if arr[m] < x:
l = m + 1
else:
r = m
r = min(n-1, l+k-1)
l = max(0, l-k)
while r-l >= k:
if x-arr[l] <= arr[r]-x:
r -= 1
else:
l += 1
return arr[l:l+k]
二分(python)
class Solution:
def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
n = len(arr)
l, r = 0, n-k
while l < r:
m = (l + r) // 2
if x-arr[m] > arr[m+k]-x:
l = m + 1
else:
r = m
return arr[l:l+k]