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Unformatted text preview: time will be bn. e. COUNTINGSORT(A, B, k) 1 for i ← 0 to k 2 do C[i] ← 0 3 for j ← 1 to length[A] 4 do C[A[j]] ← C[A[j]] + 1 5 ▹ C[i] now contains the number of elements equal to i. 6 for i ← 1 to k 7 do C[i] ← C[i] + C[i  1] 8 ▹ C[i] now contains the number of elements less than or equalto i. 9 for j ← length[A] downto 1 10 do Repeatedly move the displaced element (b = A[j]) to its sorted position (k). 11 b ← A[j] 12 k ← C[b] 13 C[b] ← C[b] − 1 14 A[j] ← a Now b must be moved to its sorted position 15 a ← b 16 j ← k 17 A[j] ← a...
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 Spring '09
 Dr.Ayman
 Algorithms, Sort, bucket sort, sorted position

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