LeetCode - Algorithms - 1470. Shuffle the Array

I can solve some easy problem on leetcode directly without IDE now. A little bit better.

Problem

1470. Shuffle the Array

Java

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class Solution {
    public int[] shuffle(int[] nums, int n) {
        int[] r = new int[2 * n];
        for (int i = 0; i < 2 * n; i++)
            r[i] = (i & 1) == 1 ? nums[n + i / 2] : nums[i / 2];
        return r;
    }
}

Submission Detail

  • 53 / 53 test cases passed.
  • Runtime: 0 ms, faster than 100.00% of Java online submissions for Shuffle the Array.
  • Memory Usage: 39.3 MB, less than 95.59% of Java online submissions for Shuffle the Array.

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class Solution {
    public int[] shuffle(int[] nums, int n) {
        int[] r = new int[2 * n];
        for (int i = 0; i < 2 * n; i++) {
            if ((i & 1) == 1) {
                r[i] = nums[n + i / 2];
            } else {
                r[i] = nums[i / 2];
            }
        }
        return r;
    }
}

Submission Detail

  • 53 / 53 test cases passed.
  • Runtime: 0 ms
  • Memory Usage: 39.7 MB, Your memory usage beats 47.43 % of java submissions.

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class Solution {
    public int[] shuffle(int[] nums, int n) {
        int[] r = new int[2 * n];
        for (int i = 0; i < 2 * n; i++)
            r[i] = nums[(i & 1) * n + i / 2];
        return r;
    }
}

Submission Detail

  • 53 / 53 test cases passed.
  • Runtime: 0 ms, faster than 100.00% of Java online submissions for Shuffle the Array.
  • Memory Usage: 39.6 MB, less than 56.99% of Java online submissions for Shuffle the Array.

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class Solution {
    public int[] shuffle(int[] nums, int n) {
        int[] r = new int[2 * n];
        for (int i = 0; i < 2 * n; i++)
            r[i] = nums[((i & 1) == 1 ? n : 0) + i / 2];
        return r;
    }
}

Submission Detail

  • 53 / 53 test cases passed.
  • Runtime: 0 ms, faster than 100.00% of Java online submissions for Shuffle the Array.
  • Memory Usage: 39.2 MB, less than 95.96% of Java online submissions for Shuffle the Array.