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Bubbling sorting principle

① compares adjacent elements, and if the previous element is larger than the latter, swap the positions of the two elements

② cycles through the above steps for each pair of adjacent elements, and the final last element is the maximum value.

Bubble sort API Design Class name Bubble Construction method Bubble: create Bubble object member method

1.public static void sort (Comparable [] a): sorts elements in an array

2.private static void greater (Comparable v comparison w); determine whether v is greater than w

3.private static void exchange (Comparable [] a drawing int x int y): swap the values at index x and index y in array a

The code for bubble sorting realizes that public class Bubble {/ / sort array a public static void sort (Comparable [] a) {for (int itima.futhmuri 1 Comparable I > 0 Comparable iMub -) {for (array element x and y exchange position private static void exchange (Comparable [] a mai int x int y) {Comparable tinca [x]; a [x] = a [y] A [y] = t;}} / / Test code class Test {public static void main (String [] args) {Integer [] a = {4, 5, 6, 3, 2, 1}; Bubble.sort (a); System.out.println (Arrays.toString (a));}}

Test results:

Analysis of time complexity of Bubble sorting

Although bubble sorting uses double-layer for loop traversal, the code that really completes the sorting is in the inner loop, so you can mainly analyze the execution times of the inner loop body.

In a worst-case scenario. That is to say, the reverse order of the array is {6 ~ 5 ~ 4 ~ 3 ~ 2 ~ 1}.

The number of comparisons of elements is: (NMU1) + (NMU2) + (NMUL3) +. + 2cm 1 =

((Nmur1) + 1) * (NMur1) / 2 = N ^ 2 / 2-N/2

The number of exchanges of elements is: (NMU1) + (NMUL2) + (NMUL3) +. + 2cm 1 =

((Nmur1) + 1) * (NMur1) / 2 = N ^ 2 / 2-N/2

The total number of execution is: 2 * (N ^ 2 / 2-N/2) = N ^ 2-N

According to the big O derivation rule, the highest order term is retained, that is, the time complexity of bubbling sorting is O (N ^ 2).

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