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Java has zero-based arrays. Fix off-by-one bug.

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edited Feb 28 at 7:02

9.2k
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If the filter does not need to be stable, you can use the following in-place algorithm:

Repeatedly find the index i of the first non-even element and the index j of the last even element. While i < j, swap the elements and repeat.

Truncate the array to j + 1 elements.

If the filter does not need to be stable, you can use the following in-place algorithm:

Repeatedly find the index i of the first non-even element and the index j of the last even element. While i < j, swap the elements and repeat.

Truncate the array to j elements.

If the filter does not need to be stable, you can use the following in-place algorithm:

Repeatedly find the index i of the first non-even element and the index j of the last even element. While i < j, swap the elements and repeat.

Truncate the array to j + 1 elements.

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answered Feb 27 at 11:37

Davislor

9.2k
19
39

If the filter does not need to be stable, you can use the following in-place algorithm:

Repeatedly find the index i of the first non-even element and the index j of the last even element. While i < j, swap the elements and repeat.

Truncate the array to j elements.