LeetCode 922 : Sort Array By Parity II
阿新 • • 發佈:2018-11-26
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
本題想要將陣列中的數與陣列下標的奇偶性統一,本人採用了以空間換時間的方式,將陣列中的奇數與偶數分別挑出,再依次插入。Accepted程式碼如下:
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; } }