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Commit e41ebf5

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feat: update solutions to leetcode problem: No.0905. Sort Array By Parity
1 parent 98d5bd8 commit e41ebf5

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5 files changed

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5 files changed

+154
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‎solution/0900-0999/0905.Sort Array By Parity/README.md‎

Lines changed: 58 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -28,27 +28,83 @@
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<li><code>0 &lt;= A[i] &lt;= 5000</code></li>
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</ol>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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双指针原地交换数组元素。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def sortArrayByParity(self, A: List[int]) -> List[int]:
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i, j = 0, len(A) - 1
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while i < j:
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if (A[i] & 1) > (A[j] & 1):
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A[i], A[j] = A[j], A[i]
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if A[i] & 1 == 0:
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i += 1
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if A[j] & 1 == 1:
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j -= 1
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return A
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] sortArrayByParity(int[] A) {
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int i = 0, j = A.length - 1;
65+
while (i < j) {
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if ((A[i] & 1) > (A[j] & 1)) {
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int t = A[i];
68+
A[i] = A[j];
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A[j] = t;
70+
}
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if ((A[i] & 1) == 0) {
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++i;
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}
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if ((A[j] & 1) == 1) {
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--j;
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}
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}
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return A;
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}
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}
81+
```
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### **JavaScript**
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```js
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/**
87+
* @param {number[]} A
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* @return {number[]}
89+
*/
90+
var sortArrayByParity = function (A) {
91+
let i = 0;
92+
let j = A.length - 1;
93+
while (i < j) {
94+
if ((A[i] & 1) > (A[j] & 1)) {
95+
const t = A[i];
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A[i] = A[j];
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A[j] = t;
98+
}
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if ((A[i] & 1) == 0) {
100+
++i;
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}
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if ((A[j] & 1) == 1) {
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--j;
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}
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}
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return A;
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};
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```
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### **...**

‎solution/0900-0999/0905.Sort Array By Parity/README_EN.md‎

Lines changed: 56 additions & 17 deletions
Original file line numberDiff line numberDiff line change
@@ -6,22 +6,14 @@
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77
<p>Given an array <code>A</code> of non-negative integers, return an array consisting of all the even elements of <code>A</code>, followed by all the odd elements of <code>A</code>.</p>
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9-
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<p>You may return any answer array that satisfies this condition.</p>
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<p>&nbsp;</p>
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<div>
2014

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<p><strong>Example 1:</strong></p>
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23-
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<pre>
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<strong>Input: </strong><span id="example-input-1-1">[3,1,2,4]</span>
@@ -32,39 +24,86 @@ The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
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</pre>
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<p>&nbsp;</p>
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39-
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<p><strong>Note:</strong></p>
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<ol>
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<li><code>1 &lt;= A.length &lt;= 5000</code></li>
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<li><code>0 &lt;= A[i] &lt;= 5000</code></li>
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</ol>
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</div>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
61-
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class Solution:
46+
def sortArrayByParity(self, A: List[int]) -> List[int]:
47+
i, j = 0, len(A) - 1
48+
while i < j:
49+
if (A[i] & 1) > (A[j] & 1):
50+
A[i], A[j] = A[j], A[i]
51+
if A[i] & 1 == 0:
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i += 1
53+
if A[j] & 1 == 1:
54+
j -= 1
55+
return A
6256
```
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### **Java**
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6660
```java
61+
class Solution {
62+
public int[] sortArrayByParity(int[] A) {
63+
int i = 0, j = A.length - 1;
64+
while (i < j) {
65+
if ((A[i] & 1) > (A[j] & 1)) {
66+
int t = A[i];
67+
A[i] = A[j];
68+
A[j] = t;
69+
}
70+
if ((A[i] & 1) == 0) {
71+
++i;
72+
}
73+
if ((A[j] & 1) == 1) {
74+
--j;
75+
}
76+
}
77+
return A;
78+
}
79+
}
80+
```
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82+
### **JavaScript**
83+
84+
```js
85+
/**
86+
* @param {number[]} A
87+
* @return {number[]}
88+
*/
89+
var sortArrayByParity = function (A) {
90+
let i = 0;
91+
let j = A.length - 1;
92+
while (i < j) {
93+
if ((A[i] & 1) > (A[j] & 1)) {
94+
const t = A[i];
95+
A[i] = A[j];
96+
A[j] = t;
97+
}
98+
if ((A[i] & 1) == 0) {
99+
++i;
100+
}
101+
if ((A[j] & 1) == 1) {
102+
--j;
103+
}
104+
}
105+
return A;
106+
};
68107
```
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### **...**
Lines changed: 19 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
1+
class Solution {
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public int[] sortArrayByParity(int[] A) {
3+
int i = 0, j = A.length - 1;
4+
while (i < j) {
5+
if ((A[i] & 1) > (A[j] & 1)) {
6+
int t = A[i];
7+
A[i] = A[j];
8+
A[j] = t;
9+
}
10+
if ((A[i] & 1) == 0) {
11+
++i;
12+
}
13+
if ((A[j] & 1) == 1) {
14+
--j;
15+
}
16+
}
17+
return A;
18+
}
19+
}

‎solution/0900-0999/0905.Sort Array By Parity/Solution.js‎

Lines changed: 11 additions & 43 deletions
Original file line numberDiff line numberDiff line change
@@ -2,53 +2,21 @@
22
* @param {number[]} A
33
* @return {number[]}
44
*/
5-
// 第一次的做法
65
var sortArrayByParity = function (A) {
7-
const len = A.length;
8-
if (len == 1) return A;
9-
let evenNumber = [];
10-
let oddNumber = [];
11-
for (let i = 0; i < len; i++) {
12-
if (A[i] % 2 == 0) {
13-
evenNumber.push(A[i]);
14-
}
15-
if (A[i] % 2 != 0) {
16-
oddNumber.push(A[i]);
17-
}
18-
}
19-
return evenNumber.concat(oddNumber);
20-
};
21-
// 修改第一次的代码,只使用一个数组,减少一次合并数组操作
22-
var sortArrayByParity = function (A) {
23-
const len = A.length;
24-
if (len == 1) return A;
25-
let eoNum = [];
26-
for (let i = 0; i < len; i++) {
27-
if (A[i] % 2 == 1) {
28-
eoNum.push(A[i]);
6+
let i = 0;
7+
let j = A.length - 1;
8+
while (i < j) {
9+
if ((A[i] & 1) > (A[j] & 1)) {
10+
const t = A[i];
11+
A[i] = A[j];
12+
A[j] = t;
2913
}
30-
if (A[i] %2 == 0) {
31-
eoNum.unshift(A[i]);
14+
if ((A[i] &1) == 0) {
15+
++i;
3216
}
33-
}
34-
return eoNum;
35-
};
36-
// 双指针做法,无需新数组,push和unshift操作,利用第三个变量进行交换
37-
var sortArrayByParity = function (A) {
38-
const len = A.length;
39-
if (len == 1) return A;
40-
let i = 0,
41-
j = len - 1;
42-
while (i < j) {
43-
if (A[i] % 2 == 1 && A[j] % 2 == 0) {
44-
let temp = A[j];
45-
A[j] = A[i];
46-
A[i] = temp;
47-
i++;
48-
j--;
17+
if ((A[j] & 1) == 1) {
18+
--j;
4919
}
50-
if (A[i] % 2 == 0) i++;
51-
if (A[j] % 2 == 1) j--;
5220
}
5321
return A;
5422
};
Lines changed: 10 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,11 @@
11
class Solution:
2-
def sortArrayByParity(self, A):
3-
"""
4-
:type A: List[int]
5-
:rtype: List[int]
6-
"""
7-
8-
return sorted(A, key = lambda i: i % 2)
2+
def sortArrayByParity(self, A: List[int]) -> List[int]:
3+
i, j = 0, len(A) - 1
4+
while i < j:
5+
if (A[i] & 1) > (A[j] & 1):
6+
A[i], A[j] = A[j], A[i]
7+
if A[i] & 1 == 0:
8+
i += 1
9+
if A[j] & 1 == 1:
10+
j -= 1
11+
return A

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