[LeetCode] 658. Find K Closest Elements 寻找K个最近元素
Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.
Example 1:
Input: [1,2,3,4,5], k=4, x=3 Output: [1,2,3,4]
Example 2:
Input: [1,2,3,4,5], k=4, x=-1 Output: [1,2,3,4]
Note:
- The value k is positive and will always be smaller than the length of the sorted array.
- Length of the given array is positive and will not exceed 104
- Absolute value of elements in the array and x will not exceed 104
UPDATE (2017/9/19):
The arr parameter had been changed to an array of integers (instead of a list of integers). Please reload the code definition to get the latest changes.
给一个有序数组和整数k, x，找出数组中离x最近的k个数。
解法1: 由于数组有序，所以最后找到的k个元素也一定是有序的，其实就是返回了一个长度为k的子数组，相当于在长度为n的数组中去掉n-k个数字，而且去掉的顺序肯定是从两头开始去，因为距离x最远的数字肯定在首尾出现。每次比较首尾两个数字跟x的距离，将距离大的那个数字删除，直到剩余的数组长度为k为止。
解法2: 二分法。 每次比较的是mid位置和x的距离跟mid+k跟x的距离，以这两者的大小关系来确定二分法折半的方向，最后找到最近距离子数组的起始位置。
Java:
public List<Integer> findClosestElements(int[] A, int k, int x) {
int left = 0, right = A.length - k;
while (left < right) {
int mid = (left + right) / 2;
if (x - A[mid] > A[mid + k] - x)
left = mid + 1;
else
right = mid;
}
List<Integer> res = new ArrayList<>();
for (int i = 0; i < k; i++) res.add(A[left + i]);
return res;
}
Java:
public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
int index = Collections.binarySearch(arr, x);
if(index < 0) index = -(index + 1);
int i = index - 1, j = index;
while(k-- > 0){
if(i<0 || (j<arr.size() && Math.abs(arr.get(i) - x) > Math.abs(arr.get(j) - x) ))j++;
else i--;
}
return arr.subList(i+1, j);
}
Java:
public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
int lo = 0, hi = arr.size() - k;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (x - arr.get(mid) > arr.get(mid+k) - x)
lo = mid + 1;
else
hi = mid;
}
return arr.subList(lo, lo + k);
}
Python:
def findClosestElements(self, A, k, x):
left, right = 0, len(A) - k
while left < right:
mid = (left + right) / 2
if x - A[mid] > A[mid + k] - x:
left = mid + 1
else:
right = mid
return A[left:left + k]
Python:
import bisect
class Solution(object):
def findClosestElements(self, arr, k, x):
"""
:type arr: List[int]
:type k: int
:type x: int
:rtype: List[int]
"""
i = bisect.bisect_left(arr, x)
left, right = i-1, i
while k:
if right >= len(arr) or \
(left >= 0 and abs(arr[left]-x) <= abs(arr[right]-x)):
left -= 1
else:
right += 1
k -= 1
return arr[left+1:right]
Python:
def findClosestElements(self, arr, k, x):
left = right = bisect.bisect_left(arr, x)
while right - left < k:
if left == 0: return arr[:k]
if right == len(arr): return arr[-k:]
if x - arr[left - 1] <= arr[right] - x: left -= 1
else: right += 1
return arr[left:right]
C++:
class Solution {
public:
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
vector<int> res = arr;
while (res.size() > k) {
int first = 0, last = res.size() - 1;
if (x - res.front() <= res.back() - x) {
res.pop_back();
} else {
res.erase(res.begin());
}
}
return res;
}
};
C++:
class Solution {
public:
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int left = 0, right = arr.size() - k;
while (left < right) {
int mid = left + (right - left) / 2;
if (x - arr[mid] > arr[mid + k] - x) left = mid + 1;
else right = mid;
}
return vector<int>(arr.begin() + left, arr.begin() + left + k);
}
};
C++:
vector<int> findClosestElements(vector<int>& A, int k, int x) {
int left = 0, right = A.size() - k;
while (left < right) {
int mid = (left + right) / 2;
if (x - A[mid] > A[mid + k] - x)
left = mid + 1;
else
right = mid;
}
return vector<int>(A.begin() + left, A.begin() + left + k);
}
All LeetCode Questions List 题目汇总
