Classic algorithm-bubble sort (Java)

Principle: Assigning large values of adjacent elements to the right

Idea: ①1. Compares the first element in the collection or array to the second element and assigns a larger value to the right;

2. Compare the second element to the third one, and the larger value to the right;

.......

(N-1). Compares the N-1 element with the nth element and assigns a larger value to the right;

From this, the maximum value of the element in the set is obtained and assigned to the element n;

② Repeat Operation ① get the new value of the element N-1, and get the new value of the element N-2;

......

Get the value of element 2 and get the value of element 1

Instance:

Sort for arr[5] = {3,5,2,8,1}:

First trip Sort:

3,5,2,8,1;

3,2,5,8,1;

3,2,5,8,1;

3,2,5,1,8;

Second trip sort:

2,3,5,1,8;

2,3,5,1,8;

2,3,1,5,8;

Third trip Sort:

2,3,1,5,8;

2,1,3,5,8;

Sequence Four:

1,2,3,5,8;

This shows: N numbers to sort complete, a total of N-1 sequencing, every I trip to the order of (n-i) times, so you can use the double loop statement, the outer control loop how many times, the inner layer control each trip cycle times, namely:

for (i = 0; i<arr.length-1;i++) {//outer control how many trips to sort

for (j = 0; j<arr.length-1-i;j++) {//The inner layer controls how many times the adjacent element size is judged by the order of each trip

if (Arr[j]>arr[j+1]) {

temp = Arr[j];

ARR[J] = arr[j+1];

ARR[J+1] = temp;

}

}

}

Classic algorithm-bubble sort (Java)