题目来源:https://leetcode.com/problems/sort-array-by-parity/
LeetCode官方题解及解析:905. Sort Array By Parity
题目
Given an array A
of non-negative integers, return an array consisting of all the even elements of A
, followed by all the odd elements of A
.
You may return any answer array that satisfies this condition.
Example 1:
Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Note:
1 <= A.length <= 5000
0 <= A[i] <= 5000
解答1[Java]:二次筛选
class Solution {
public int[] sortArrayByParity(int[] A) {
int[] result = new int[A.length];
int j = 0;
for(int i = 0; i < A.length; i++)
{
if(A[i]%2 == 0)
result[j++] = A[i];
}
for(int i = 0; i < A.length; i++)
{
if(A[i]%2 == 1)
result[j++] = A[i];
}
return result;
}
}
复杂度分析
- Time Complexity: , where is the length of
A
. - Space Complexity: for the sort, depending on the built-in implementation of
sort
.
解答2[Java]:使用Comparator()
class Solution {
public int[] sortArrayByParity(int[] A) {
Integer[] B = new Integer[A.length];
for (int t = 0; t < A.length; ++t)
B[t] = A[t];
Arrays.sort(B, (a, b) -> Integer.compare(a%2, b%2));
for (int t = 0; t < A.length; ++t)
A[t] = B[t];
return A;
/* Alternative:
return Arrays.stream(A)
.boxed()
.sorted((a, b) -> Integer.compare(a%2, b%2))
.mapToInt(i -> i)
.toArray();
*/
}
}
因为 Comparator
不能对原生类型使用,所以要对原生数组进行装箱,把 Int[]
改为 Integer[]
。排序完成之后再转换为原生数组。
时间复杂度
- Time Complexity: , where is the length of
A
. - Space Complexity: for the sort, depending on the built-in implementation of
sort
.
解答3[Java]:使用快排
class Solution {
public int[] sortArrayByParity(int[] A) {
int i = 0, j = A.length - 1;
while (i < j) {
if (A[i] % 2 > A[j] % 2) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
if (A[i] % 2 == 0) i++;
if (A[j] % 2 == 1) j--;
}
return A;
}
}
思路
We'll maintain two pointers i
and j
. The loop invariant is everything below i
has parity 0
(ie. A[k] % 2 == 0
when k < i
), and everything above j
has parity 1
.
Then, there are 4 cases for (A[i] % 2, A[j] % 2)
:
- If it is
(0, 1)
, then everything is correct:i++
andj--
. - If it is
(1, 0)
, we swap them so they are correct, then continue. - If it is
(0, 0)
, only thei
place is correct, so wei++
and continue. - If it is
(1, 1)
, only thej
place is correct, so wej--
and continue.
Throughout all 4 cases, the loop invariant is maintained, and j-i
is getting smaller. So eventually we will be done with the array sorted as desired.
时间复杂度
- Time Complexity: , where is the length of
A
. Each step of the while loop makesj-i
decrease by at least one. (Note that while quicksort is normally, this is because we only need one pass to sort the elements.) - Space Complexity: in additional space complexity.