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feat: update solutions to lc problem: No.2529 (#2553)
No.2529.Maximum Count of Positive Integer and Negative Integer
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‎solution/2500-2599/2529.Maximum Count of Positive Integer and Negative Integer/README.md

+65-91
Original file line numberDiff line numberDiff line change
@@ -60,29 +60,28 @@
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### 方法一：遍历
6262

63-
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。
63+
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。
6464

65-
时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
65+
时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
6666

6767
<!-- tabs:start -->
6868

6969
```python
7070
class Solution:
7171
def maximumCount(self, nums: List[int]) -> int:
72-
a = sum(v > 0 for v in nums)
73-
b = sum(v < 0 for v in nums)
72+
a = sum(x > 0 for x in nums)
73+
b = sum(x < 0 for x in nums)
7474
return max(a, b)
7575
```
7676

7777
```java
7878
class Solution {
7979
public int maximumCount(int[] nums) {
8080
int a = 0, b = 0;
81-
for (int v : nums) {
82-
if (v > 0) {
81+
for (int x : nums) {
82+
if (x > 0) {
8383
++a;
84-
}
85-
if (v < 0) {
84+
} else if (x < 0) {
8685
++b;
8786
}
8887
}
@@ -96,11 +95,10 @@ class Solution {
9695
public:
9796
int maximumCount(vector<int>& nums) {
9897
int a = 0, b = 0;
99-
for (int& v : nums) {
100-
if (v > 0) {
98+
for (int x : nums) {
99+
if (x > 0) {
101100
++a;
102-
}
103-
if (v < 0) {
101+
} else if (x < 0) {
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++b;
105103
}
106104
}
@@ -111,12 +109,11 @@ public:
111109
112110
```go
113111
func maximumCount(nums []int) int {
114-
a, b := 0, 0
115-
for _, v := range nums {
116-
if v > 0 {
112+
var a, b int
113+
for _, x := range nums {
114+
if x > 0 {
117115
a++
118-
}
119-
if v < 0 {
116+
} else if x < 0 {
120117
b++
121118
}
122119
}
@@ -126,47 +123,50 @@ func maximumCount(nums []int) int {
126123

127124
```ts
128125
function maximumCount(nums: number[]): number {
129-
const count = [0, 0];
130-
for (const num of nums) {
131-
if (num < 0) {
132-
count[0]++;
133-
} else if (num > 0) {
134-
count[1]++;
126+
let [a, b] = [0, 0];
127+
for (const x of nums) {
128+
if (x > 0) {
129+
++a;
130+
} else if (x < 0) {
131+
++b;
135132
}
136133
}
137-
return Math.max(...count);
134+
return Math.max(a, b);
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}
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```
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141138
```rust
142139
impl Solution {
143140
pub fn maximum_count(nums: Vec<i32>) -> i32 {
144-
let mut count = [0, 0];
145-
for &num in nums.iter() {
146-
if num < 0 {
147-
count[0] += 1;
148-
} else if num > 0 {
149-
count[1] += 1;
141+
let mut a = 0;
142+
let mut b = 0;
143+
144+
for x in nums {
145+
if x > 0 {
146+
a += 1;
147+
} else if x < 0 {
148+
b += 1;
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}
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}
152-
*count.iter().max().unwrap()
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152+
std::cmp::max(a, b)
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}
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}
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```
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157157
```c
158-
#define max(a, b) (((a) > (b)) ? (a) : (b))
158+
#define max(a, b) (a > b ? a : b)
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160160
int maximumCount(int* nums, int numsSize) {
161-
int count[2] = {0};
162-
for (int i = 0; i < numsSize; i++) {
163-
if (nums[i] < 0) {
164-
count[0]++;
165-
} else if (nums[i] > 0) {
166-
count[1]++;
161+
int a = 0, b = 0;
162+
for (int i = 0; i < numsSize; ++i) {
163+
if (nums[i] > 0) {
164+
++a;
165+
} else if (nums[i] < 0) {
166+
++b;
167167
}
168168
}
169-
return max(count[0], count[1]);
169+
return max(a, b);
170170
}
171171
```
172172
@@ -176,7 +176,7 @@ int maximumCount(int* nums, int numsSize) {
176176
177177
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。
178178
179-
时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
179+
时间复杂度 $O(\log n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
180180
181181
<!-- tabs:start -->
182182
@@ -232,97 +232,71 @@ func maximumCount(nums []int) int {
232232

233233
```ts
234234
function maximumCount(nums: number[]): number {
235-
const search = (target: number) => {
236-
let left = 0;
237-
let right = n;
235+
const search = (x: number): number => {
236+
let [left, right] = [0, nums.length];
238237
while (left < right) {
239-
const mid = (left + right) >>> 1;
240-
if (nums[mid] < target) {
241-
left = mid + 1;
242-
} else {
238+
const mid = (left + right) >> 1;
239+
if (nums[mid] >= x) {
243240
right = mid;
241+
} else {
242+
left = mid + 1;
244243
}
245244
}
246245
return left;
247246
};
248-
const n = nums.length;
249-
const i = search(0);
250-
const j = search(1);
251-
return Math.max(i, n - j);
247+
const i = search(1);
248+
const j = search(0);
249+
const [a, b] = [nums.length - i, j];
250+
return Math.max(a, b);
252251
}
253252
```
254253

255254
```rust
256255
impl Solution {
257-
fn search(nums: &Vec<i32>, target: i32) -> usize {
256+
fn search(nums: &Vec<i32>, x: i32) -> usize {
258257
let mut left = 0;
259258
let mut right = nums.len();
260259
while left < right {
261260
let mid = (left + right) >> 1;
262-
if nums[mid] < target {
263-
left = mid + 1;
264-
} else {
261+
if nums[mid] >= x {
265262
right = mid;
263+
} else {
264+
left = mid + 1;
266265
}
267266
}
268267
left
269268
}
270269

271270
pub fn maximum_count(nums: Vec<i32>) -> i32 {
272271
let n = nums.len();
273-
let i = Self::search(&nums, 0);
274-
let j = Self::search(&nums, 1);
275-
i.max(n - j) as i32
272+
let i = Self::search(&nums, 1);
273+
let j = Self::search(&nums, 0);
274+
(n - i).max(j) as i32
276275
}
277276
}
278277
```
279278

280279
```c
281-
#define max(a, b) (((a) > (b)) ? (a) : (b))
280+
#define max(a, b) (a > b ? a : b)
282281

283-
int search(int* nums, int numsSize, int target) {
282+
int search(int* nums, int numsSize, int x) {
284283
int left = 0;
285284
int right = numsSize;
286285
while (left < right) {
287286
int mid = (left + right) >> 1;
288-
if (nums[mid] < target) {
289-
left = mid + 1;
290-
} else {
287+
if (nums[mid] >= x) {
291288
right = mid;
289+
} else {
290+
left = mid + 1;
292291
}
293292
}
294293
return left;
295294
}
296295

297296
int maximumCount(int* nums, int numsSize) {
298-
int i = search(nums, numsSize, 0);
299-
int j = search(nums, numsSize, 1);
300-
return max(i, numsSize - j);
301-
}
302-
```
303-
304-
<!-- tabs:end -->
305-
306-
### 方法三
307-
308-
<!-- tabs:start -->
309-
310-
```rust
311-
impl Solution {
312-
pub fn maximum_count(nums: Vec<i32>) -> i32 {
313-
let mut a = 0;
314-
let mut b = 0;
315-
316-
for n in nums {
317-
if n > 0 {
318-
a += 1;
319-
} else if n < 0 {
320-
b += 1;
321-
}
322-
}
323-
324-
std::cmp::max(a, b)
325-
}
297+
int i = search(nums, numsSize, 1);
298+
int j = search(nums, numsSize, 0);
299+
return max(numsSize - i, j);
326300
}
327301
```
328302

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