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@@ -52,6 +52,26 @@
<!-- 这里可写通用的实现逻辑 -->
**方法一:枚举**
我们在 $[0, n)$ 的范围内枚举下标 $i,ドル对于每个下标 $i,ドル我们在 $[0, n)$ 的范围内枚举下标 $j,ドル如果 $|i - j| \leq k$ 且 $nums[j] == key,ドル那么 $i$ 就是一个 K 近邻下标,我们将 $i$ 加入答案数组中,然后跳出内层循环,枚举下一个下标 $i$。
时间复杂度 $O(n^2),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
**方法二:预处理 + 二分查找**
我们可以预处理得到所有等于 $key$ 的元素的下标,记录在数组 $idx$ 中。数组 $idx$ 中的所有下标元素是按照升序排列的,
接下来,我们枚举下标 $i,ドル对于每个下标 $i,ドル我们可以使用二分查找的方法在数组 $idx$ 中查找 $[i - k, i + k]$ 范围内的元素,如果存在元素,那么 $i$ 就是一个 K 近邻下标,我们将 $i$ 加入答案数组中。
时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
**方法三:双指针**
我们枚举下标 $i,ドル用一个指针 $j$ 指向满足 $j \geq i - k$ 且 $nums[j] = key$ 的最小下标,如果 $j$ 存在且 $j \leq i + k,ドル那么 $i$ 就是一个 K 近邻下标,我们将 $i$ 加入答案数组中。
时间复杂度 $O(n),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
<!-- tabs:start -->
### **Python3**
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@@ -64,10 +84,34 @@ class Solution:
ans = []
n = len(nums)
for i in range(n):
for j in range(n):
if abs(i - j) <= k and nums[j] == key:
ans.append(i)
break
if any(abs(i - j) <= k and nums[j] == key for j in range(n)):
ans.append(i)
return ans
```
```python
class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
idx = [i for i, x in enumerate(nums) if x == key]
ans = []
for i in range(len(nums)):
l = bisect_left(idx, i - k)
r = bisect_right(idx, i + k) - 1
if l <= r:
ans.append(i)
return ans
```
```python
class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
ans = []
j, n = 0, len(nums)
for i in range(n):
while j < i - k or (j < n and nums[j] != key):
j += 1
if j < n and j <= (i + k):
ans.append(i)
return ans
```
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@@ -93,22 +137,45 @@ class Solution {
}
```
### **TypeScript**
```java
class Solution {
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
List<Integer> idx = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] == key) {
idx.add(i);
}
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
int l = Collections.binarySearch(idx, i - k);
int r = Collections.binarySearch(idx, i + k + 1);
l = l < 0 ? -l - 1 : l;
r = r < 0 ? -r - 2 : r - 1;
if (l <= r) {
ans.add(i);
}
}
return ans;
}
}
```
```ts
function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
let ans = [];
for (let j = 0; j < n; j++) {
if (nums[j] == key) {
for (let i = j - k; i <= j + k; i++) {
if (i >= 0 && i < n && !ans.includes(i)) {
ans.push(i);
}
```java
class Solution {
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
int n = nums.length;
List<Integer> ans = new ArrayList<>();
for (int i = 0, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] != key)) {
++j;
}
if (j < n && j <= i + k) {
ans.add(i);
}
}
return ans;
}
return ans;
}
```
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@@ -133,15 +200,56 @@ public:
};
```
```cpp
class Solution {
public:
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
vector<int> idx;
int n = nums.size();
for (int i = 0; i < n; ++i) {
if (nums[i] == key) {
idx.push_back(i);
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
auto it1 = lower_bound(idx.begin(), idx.end(), i - k);
auto it2 = upper_bound(idx.begin(), idx.end(), i + k) - 1;
if (it1 <= it2) {
ans.push_back(i);
}
}
return ans;
}
};
```
```cpp
class Solution {
public:
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
int n = nums.size();
vector<int> ans;
for (int i = 0, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] != key)) {
++j;
}
if (j < n && j <= i + k) {
ans.push_back(i);
}
}
return ans;
}
};
```
### **Go**
```go
func findKDistantIndices(nums []int, key int, k int) []int {
n := len(nums)
var ans []int
for i := 0; i < n; i++ {
for j, v := range nums {
if abs(i-j) <= k && v == key {
func findKDistantIndices(nums []int, key int, k int) (ans []int) {
for i := range nums {
for j, x := range nums {
if abs(i-j) <= k && x == key {
ans = append(ans, i)
break
}
Expand All
@@ -158,6 +266,107 @@ func abs(x int) int {
}
```
```go
func findKDistantIndices(nums []int, key int, k int) (ans []int) {
idx := []int{}
for i, x := range nums {
if x == key {
idx = append(idx, i)
}
}
for i := range nums {
l := sort.SearchInts(idx, i-k)
r := sort.SearchInts(idx, i+k+1) - 1
if l <= r {
ans = append(ans, i)
}
}
return
}
```
```go
func findKDistantIndices(nums []int, key int, k int) (ans []int) {
n := len(nums)
for i, j := 0, 0; i < n; i++ {
for j < i-k || (j < n && nums[j] != key) {
j++
}
if j < n && j <= i+k {
ans = append(ans, i)
}
}
return
}
```
### **TypeScript**
```ts
function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (Math.abs(i - j) <= k && nums[j] === key) {
ans.push(i);
break;
}
}
}
return ans;
}
```
```ts
function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const idx: number[] = [];
for (let i = 0; i < n; i++) {
if (nums[i] === key) {
idx.push(i);
}
}
const search = (x: number): number => {
let [l, r] = [0, idx.length];
while (l < r) {
const mid = (l + r) >> 1;
if (idx[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
const l = search(i - k);
const r = search(i + k + 1) - 1;
if (l <= r) {
ans.push(i);
}
}
return ans;
}
```
```ts
function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] !== key)) {
++j;
}
if (j < n && j <= i + k) {
ans.push(i);
}
}
return ans;
}
```
### **...**
```
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