版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/wuqingdeqing/article/details/84534133
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 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
本题想要将数组中的数与数组下标的奇偶性统一,本人采用了以空间换时间的方式,将数组中的奇数与偶数分别挑出,再依次插入。Accepted代码如下:
class Solution {
public int[] sortArrayByParityII(int[] A) {
int l = A.length;
int[] B = new int[l / 2];
int[] C = new int[l / 2];
for (int i = 0, j = 0, k = 0; i < A.length; i++) {
if (A[i] % 2 == 0) {
B[j++] = A[i];
} else {
C[k++] = A[i];
}
}
for (int i = 0, j = 0, k = 0; i < A.length; i++) {
if (i % 2 == 0) {
A[i] = B[j++];
} else {
A[i] = C[k++];
}
}
return A;
}
}