922 Sort Array By Parity II 按奇偶排序数组 II
Description:
Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.
Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.
You may return any answer array that satisfies this condition.
Example:
Example 1:
Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Note:
2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000
题目描述:
给定一个非负整数数组 A， A 中一半整数是奇数，一半整数是偶数。
对数组进行排序，以便当 A[i] 为奇数时，i 也是奇数；当 A[i] 为偶数时， i 也是偶数。
你可以返回任何满足上述条件的数组作为答案。
示例 :
输入：[4,2,5,7]
输出：[4,5,2,7]
解释：[4,7,2,5]，[2,5,4,7]，[2,7,4,5] 也会被接受。
提示：
2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000
思路:
双指针法
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
vector<int> sortArrayByParityII(vector<int>& A)
{
int i = 0, j = 1;
while (i < A.size())
{
if ((A[i] & 1))
{
while ((A[j] & 1)) j += 2;
A[i] ^= A[j];
A[j] ^= A[i];
A[i] ^= A[j];
}
i += 2;
}
return A;
}
};
Java:
class Solution {
public int[] sortArrayByParityII(int[] A) {
int i = 0, j = 1;
while (i < A.length) {
if ((A[i] & 1) != 0) {
while ((A[j] & 1) != 0) j += 2;
A[i] ^= A[j];
A[j] ^= A[i];
A[i] ^= A[j];
}
i += 2;
}
return A;
}
}
Python:
class Solution:
def sortArrayByParityII(self, A: List[int]) -> List[int]:
return [i for pair in zip([i for i in A if not (i & 1)], [i for i in A if (i & 1)]) for i in pair]
