LeetCode - Algorithms - 905. Sort Array By Parity

It’s easy indeed.

Problem

905. Sort Array By Parity

Java

Two Pass

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class Solution {
    public int[] sortArrayByParity(int[] A) {
        final int N = A.length;
        int[] B = new int[N];
        int left=0, right=N-1;
        for(int i=0; i<N; i++) {
            if ((A[i]&1)==0) {
                B[left++]=A[i];
            }
            else {
                B[right--]=A[i];
            }
        }
        return B;
    }
}

Submission Detail

  • 285 / 285 test cases passed.
  • Runtime: 1 ms, faster than 99.47% of Java online submissions for Sort Array By Parity.
  • Memory Usage: 40.1 MB, less than 30.13% of Java online submissions for Sort Array By Parity.

In-Place

© Solution - Approach 3: In-Place

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class Solution {
    public int[] sortArrayByParity(int[] A) {
        final int N = A.length;
        int left = 0, right = N - 1;
        int temp;
        while (left < right) {
            if ((A[left] & 1)==1 && (A[right] & 1)==0) {
                temp = A[left];
                A[left] = A[right];
                A[right] = temp;
            }
            if ((A[left] & 1) == 0) left++;
            if ((A[right] & 1) == 1) right--;
        }
        return A;
    }
}

Submission Detail

  • 285 / 285 test cases passed.
  • Runtime: 0 ms, faster than 100.00% of Java online submissions for Sort Array By Parity.
  • Memory Usage: 39.6 MB, less than 30.13% of Java online submissions for Sort Array By Parity.