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feat: update solutions to lc problem: No.2529 (#2553)
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No.2529.Maximum Count of Positive Integer and Negative Integer
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@yanglbme
yanglbme authored Apr 8, 2024
1 parent ea860e7 commit 96852d1
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156 changes: 65 additions & 91 deletions ...2500-2599/2529.Maximum Count of Positive Integer and Negative Integer/README.md
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Expand Up @@ -60,29 +60,28 @@

### 方法一：遍历

遍历数组，统计正整数和负整数的个数 $a$ 和 $b$，返回 $a$ 和 $b$ 中的较大值即可。
我们可以直接遍历数组，统计正整数和负整数的个数 $a$ 和 $b$，返回 $a$ 和 $b$ 中的较大值即可。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

```python
class Solution:
def maximumCount(self, nums: List[int]) -> int:
a = sum(v > 0 for v in nums)
b = sum(v < 0 for v in nums)
a = sum(x > 0 for x in nums)
b = sum(x < 0 for x in nums)
return max(a, b)
```

```java
class Solution {
public int maximumCount(int[] nums) {
int a = 0, b = 0;
for (int v : nums) {
if (v > 0) {
for (int x : nums) {
if (x > 0) {
++a;
}
if (v < 0) {
} else if (x < 0) {
++b;
}
}
Expand All @@ -96,11 +95,10 @@ class Solution {
public:
int maximumCount(vector<int>& nums) {
int a = 0, b = 0;
for (int& v : nums) {
if (v > 0) {
for (int x : nums) {
if (x > 0) {
++a;
}
if (v < 0) {
} else if (x < 0) {
++b;
}
}
Expand All @@ -111,12 +109,11 @@ public:
```go
func maximumCount(nums []int) int {
a, b := 0, 0
for _, v := range nums {
if v > 0 {
var a, b int
for _, x := range nums {
if x > 0 {
a++
}
if v < 0 {
} else if x < 0 {
b++
}
}
Expand All @@ -126,47 +123,50 @@ func maximumCount(nums []int) int {

```ts
function maximumCount(nums: number[]): number {
const count = [0, 0];
for (const num of nums) {
if (num < 0) {
count[0]++;
} else if (num > 0) {
count[1]++;
let [a, b] = [0, 0];
for (const x of nums) {
if (x > 0) {
++a;
} else if (x < 0) {
++b;
}
}
return Math.max(...count);
return Math.max(a, b);
}
```

```rust
impl Solution {
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let mut count = [0, 0];
for &num in nums.iter() {
if num < 0 {
count[0] += 1;
} else if num > 0 {
count[1] += 1;
let mut a = 0;
let mut b = 0;

for x in nums {
if x > 0 {
a += 1;
} else if x < 0 {
b += 1;
}
}
*count.iter().max().unwrap()

std::cmp::max(a, b)
}
}
```

```c
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define max(a, b) (a > b ? a : b)

int maximumCount(int* nums, int numsSize) {
int count[2] = {0};
for (int i = 0; i < numsSize; i++) {
if (nums[i] < 0) {
count[0]++;
} else if (nums[i] > 0) {
count[1]++;
int a = 0, b = 0;
for (int i = 0; i < numsSize; ++i) {
if (nums[i] > 0) {
++a;
} else if (nums[i] < 0) {
++b;
}
}
return max(count[0], count[1]);
return max(a, b);
}
```
Expand All @@ -176,7 +176,7 @@ int maximumCount(int* nums, int numsSize) {
由于数组是按非递减顺序排列的，因此可以使用二分查找找到第一个大于等于 $1$ 的元素的下标 $i$ 以及第一个大于等于 $0$ 的元素的下标 $j$，那么正整数的个数 $a = n - i$，负整数的个数 $b = j$，返回 $a$ 和 $b$ 中的较大值即可。
时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
时间复杂度 $O(\log n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
<!-- tabs:start -->
Expand Down Expand Up @@ -232,97 +232,71 @@ func maximumCount(nums []int) int {

```ts
function maximumCount(nums: number[]): number {
const search = (target: number) => {
let left = 0;
let right = n;
const search = (x: number): number => {
let [left, right] = [0, nums.length];
while (left < right) {
const mid = (left + right) >>> 1;
if (nums[mid] < target) {
left = mid + 1;
} else {
const mid = (left + right) >> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};
const n = nums.length;
const i = search(0);
const j = search(1);
return Math.max(i, n - j);
const i = search(1);
const j = search(0);
const [a, b] = [nums.length - i, j];
return Math.max(a, b);
}
```

```rust
impl Solution {
fn search(nums: &Vec<i32>, target: i32) -> usize {
fn search(nums: &Vec<i32>, x: i32) -> usize {
let mut left = 0;
let mut right = nums.len();
while left < right {
let mid = (left + right) >> 1;
if nums[mid] < target {
left = mid + 1;
} else {
if nums[mid] >= x {
right = mid;
} else {
left = mid + 1;
}
}
left
}

pub fn maximum_count(nums: Vec<i32>) -> i32 {
let n = nums.len();
let i = Self::search(&nums, 0);
let j = Self::search(&nums, 1);
i.max(n - j) as i32
let i = Self::search(&nums, 1);
let j = Self::search(&nums, 0);
(n - i).max(j) as i32
}
}
```

```c
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define max(a, b) (a > b ? a : b)

int search(int* nums, int numsSize, int target) {
int search(int* nums, int numsSize, int x) {
int left = 0;
int right = numsSize;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] < target) {
left = mid + 1;
} else {
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}

int maximumCount(int* nums, int numsSize) {
int i = search(nums, numsSize, 0);
int j = search(nums, numsSize, 1);
return max(i, numsSize - j);
}
```
<!-- tabs:end -->
### 方法三
<!-- tabs:start -->
```rust
impl Solution {
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let mut a = 0;
let mut b = 0;
for n in nums {
if n > 0 {
a += 1;
} else if n < 0 {
b += 1;
}
}
std::cmp::max(a, b)
}
int i = search(nums, numsSize, 1);
int j = search(nums, numsSize, 0);
return max(numsSize - i, j);
}
```
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