1. Bubble sort
Bubble sort, the array is arranged from small to large, and adjacent elements are compared. If the first one is bigger than the second, swap them two.
Example:
/*
* 冒泡排序
* 前后两个元素相比较，最大的往后排，每一轮都可以找到一个最大的元素
*/
publicstaticint[] bubbleSort(int[] nums){
int temp =0;
for (int i = 0; i < nums.length-1; i++) {
for (int j = 0; j < nums.length-1-i; j++) {
if (nums[j]>nums[j+1]){
temp= nums[j];
nums[j]=nums[j+1];
nums[j+1]= temp;
}
}
}
return nums;
}Before sorting: {3,1,4,9,5,10,6,19,0,7,2,3}
After sorting: [0,1, 2, 3, 3, 4, 5, 6, 7, 9, 10, 19]
Time complexity: O(n 2 )
2. Quick sort
Quick sorting is to divide the records to be sorted into two independent parts by sorting. One part of the records has a smaller keyword than the other part. Then the two parts are sorted separately until the entire sequence is in order.
Example:
/*
* 快速排序
* 取第一个元素为基数，循环比较最小位置值与最大位置值，较大的往后排，较小的往前排，
* 然后并较小的位数加一，较大的位数减一，循环结果就是中间位数，
* 最后就是以中间两部分分开做递归运算再次比较排列，直到正确排序为止
*/
publicstaticvoid quickSort(int[] nums, int low, int high) {
if (low < high) {
int temp = nums[low]; // 选定的基准值（第一个数值作为基准值）
int middle; // 记录临时中间值
int i = low, j = high;
do {
while ((nums[i] < temp)&& (i < high))
i++;
while ((nums[j] > temp)&& (j > low))
j--;
if (i <= j) {
middle= nums[i];
nums[i]= nums[j];
nums[j]= middle;
i++;
j--;
}
}while(i <= j);
if (low < j)
quickSort(nums,low, j);
if (high > i)
quickSort(nums,i, high);
}
}Before sorting: {3,1,4,9,5,10,6,19,0,7,2,3}
After sorting: [0,1, 2, 3, 3, 4, 5, 6, 7, 9, 10, 19]
Time complexity: O(n 2 )
3. Select sort
Select sorting, each time find the smallest value in the sequence, and then put it in the first position.
Example:
/*
* 选择排序
* 从后往前排，设默认最小元素与较大位置比较，如果比最小元素大，
* 则将该位置记为最小元素，并且循环比较最小元素，找到最小元素，
* 最后将最小元素赋值最前方，
*/
publicstaticvoid selectSort(int[] nums) {
int size = nums.length, temp;
for (int i = 0; i < size; i++) {
int k = i; //临时最小元素位数
for (int j = size - 1; j >i; j--) {
if (nums[j] < nums[k]) k = j;
}
temp = nums[i];
nums[i] = nums[k];
nums[k] = temp;
}
} Before sorting: {3,1,4,9,5,10,6,19,0,7,2,3}
After sorting: [0,1, 2, 3, 3, 4, 5, 6, 7, 9, 10, 19]
Time complexity: O(n 2 )
4. Insertion sort
Insertion sort, by constructing an ordered sequence, for unsorted data, scan from back to front in the sorted sequence, find the corresponding position and insert.
Example:
/*
* 插入排序
* 基准元素相邻元素比较，较大的往后排，然后从较大的位置开始往后与基准元素比较排列，
* 最后把最小的插入较小位
*/
publicstaticvoid insertSort(int[] nums) {
int size = nums.length, temp, j;
for(int i=1; i<size; i++) {
temp = nums[i];//基准元素
for(j = i; j > 0 && temp < nums[j-1];j--)
nums[j] = nums[j-1];
nums[j] = temp;
}
}Before sorting: {3,1,4,9,5,10,6,19,0,7,2,3}
After sorting: [0,1, 2, 3, 3, 4, 5, 6, 7, 9, 10, 19]
Time complexity: O(n 2 )