算法 | Java 常见排序算法(纯代码)
目录
汇总
序号 | 排序算法 | 平均时间 | 最好情况 | 最差情况 | 稳定度 | 额外空间 | 备注 | 相对时间 |
---|---|---|---|---|---|---|---|---|
1 | 冒泡算法 | O(n2) | O(n) | O(n2) | 稳定 | O(1) | n 越小越好 | 182 ms |
2 | 选择算法 | O(n2) | O(n2) | O(n2) | 不稳定 | O(1) | n 越小越好 | 53 ms |
3 | 插入算法 | O(n2) | O(n) | O(n2) | 稳定 | O(1) | 大部分排序好时好 | 16 ms |
4 | 快速算法 | O(nlog2n) | O(nlog2n) | O(n2) | 不稳定 | O(nlog2n) | n 大时好 | 719 ms |
5 | 归并算法 | O(nlog2n) | O(nlog2n) | O(nlog2n) | 稳定 | O(n) | n 大时好 | 550 ms |
6 | 希尔算法 | O(nlog2n) | O(n) | O(n2) | 不稳定 | O(1) | 197 ms/4 ms | |
7 | 堆排序 | O(nlog2n) | O(nlog2n) | O(nlog2n) | 不稳定 | O(1) | n 大时好 | 3 ms |
8 | 计数排序 | O(n+k) | O(n+k) | O(n+k) | 稳定 | O(n+k) | k 是桶的数量 | 2 ms |
9 | 桶排序 | O(n+k) | O(n) | O(n2) | 稳定 | O(n+k) | 11 ms | |
10 | 基数排序 | O(n*k) | O(n*k) | O(n*k) | 稳定 | O(n+k) | 4 ms | |
11 | 优先队列 | 不稳定 | O(n) | 9 ms | ||||
12 | Java API | O(1) | 4 ms |
1. 冒泡排序
每轮循环确定最值;
public void bubbleSort(int[] nums){
int temp;
boolean isSort = false; //优化,发现排序好就退出
for (int i = 0; i < nums.length-1; i++) {
for (int j = 0; j < nums.length-1-i; j++) { //每次排序后能确定较大值
if(nums[j] > nums[j+1]){
isSort = true;
temp = nums[j];
nums[j] = nums[j+1];
nums[j+1] = temp;
}
}
if(!isSort){
return;
} else {
isSort = false;
}
}
}
2. 选择排序
每次选出最值,再交换到边上;
public void selectSort(int[] nums){
for (int i = 0; i < nums.length-1; i++) {
int index = i;
int minNum = nums[i];
for (int j = i+1; j < nums.length; j++) {
if(nums[j] < minNum){
minNum = nums[j];
index = j;
}
}
if(index != i){
nums[index] = nums[i];
nums[i] = minNum;
}
}
}
3. 插入排序
对循环的每个数找到属于自己的位置插入;
public void insertionSort(int[] nums){
for (int i = 1; i < nums.length; i++) {
int j = i;
int insertNum = nums[i];
while(j-1 >= 0 && nums[j-1] > insertNum){
nums[j] = nums[j-1];
j--;
}
nums[j] = insertNum;
}
}
4. 快速排序
选一个基本值,小于它的放一边,大于它的放另一边;
public void quickSortDfs(int[] nums, int left, int right){
if(left > right){
return;
}
int l = left;
int r = right;
int baseNum = nums[left];
while(l < r){
//必须右边先走
while(nums[r] >= baseNum && l < r){
r--;
}
while(nums[l] <= baseNum && l < r){
l++;
}
int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
nums[left] = nums[l];
nums[l] = baseNum;
quickSortDfs(nums, left, r-1);
quickSortDfs(nums, l+1, right);
}
5. 归并排序
分治算法;
//归
public void mergeSortDfs(int[] nums, int l, int r){
if(l >= r){
return;
}
int m = (l+r)/2;
mergeSortDfs(nums, l, m);
mergeSortDfs(nums, m+1, r);
merge(nums, l, m, r);
}
//并
private void merge(int[] nums, int left, int mid, int right){
int[] temp = new int[right-left+1];
int l = left;
int m = mid+1;
int i = 0;
while(l <= mid && m <= right){
if(nums[l] < nums[m]){
temp[i++] = nums[l++];
} else {
temp[i++] = nums[m++];
}
}
while(l <= mid){
temp[i++] = nums[l++];
}
while(m <= right){
temp[i++] = nums[m++];
}
System.arraycopy(temp, 0, nums, left, temp.length);
}
6. 希尔排序
引入步长减少数字交换次数提高效率;
6.1 希尔-冒泡排序(慢)
public void shellBubbleSort(int[] nums){
for (int step = nums.length/2; step > 0 ; step /= 2) {
for (int i = step; i < nums.length; i++) {
for (int j = i-step; j >= 0; j -= step) {
if(nums[j] > nums[j+step]){
int temp = nums[j];
nums[j] = nums[j+step];
nums[j+step] = temp;
}
}
}
}
}
6.2 希尔-插入排序(快)
public void shellInsertSort(int[] nums){
for (int step = nums.length/2; step > 0; step /= 2) {
for (int i = step; i < nums.length; i++) {
int j = i;
int insertNum = nums[i];
while(j-step >= 0 && nums[j-step] > insertNum){
nums[j] = nums[j-step];
j-=step;
}
nums[j] = insertNum;
}
}
}
7. 堆排序
大顶堆实现升序,每次将最大值移到堆的最后一个位置上;
public void heapSort2(int[] nums) {
for(int i = nums.length/2-1; i >= 0; i--){
sift(nums, i, nums.length);
}
for (int i = nums.length-1; i > 0; i--) {
int temp = nums[0];
nums[0] = nums[i];
nums[i] = temp;
sift(nums, 0, i);
}
}
private void sift(int[] nums, int parent, int len) {
int value = nums[parent];
for (int child = 2*parent +1; child < len; child = child*2 +1) {
if(child+1 < len && nums[child+1] > nums[child]){
child++;
}
if(nums[child] > value){
nums[parent] = nums[child];
parent = child;
} else {
break;
}
}
nums[parent] = value;
}
8. 计数排序
按顺序统计每个数出现次数;
public void countSort(int[] nums){
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for(int num : nums){
max = Math.max(max, num);
min = Math.min(min, num);
}
int[] countMap = new int[max-min+1];
for(int num : nums){
countMap[num-min]++;
}
int i = 0;
int j = 0;
while(i < nums.length && j < countMap.length){
if(countMap[j] > 0){
nums[i] = j+min;
i++;
countMap[j]--;
} else {
j++;
}
}
}
9. 桶排序
类似计数排序,不同点在于统计的是某个区间(桶)里的数;
public void bucketSort(int[] nums){
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for(int num : nums){
max = Math.max(max, num);
min = Math.min(min, num);
}
int bucketCount = (max-min)/nums.length+1;
List<List<Integer>> bucketList = new ArrayList<>();
for (int i = 0; i < bucketCount; i++) {
bucketList.add(new ArrayList<>());
}
for(int num : nums){
int index = (num-min)/nums.length;
bucketList.get(index).add(num);
}
for(List<Integer> bucket : bucketList){
Collections.sort(bucket);
}
int j = 0;
for(List<Integer> bucket : bucketList){
for(int num : bucket){
nums[j] = num;
j++;
}
}
}
10. 基数排序
按个、十、百位依次归类排序;
public void radixSort(int[] nums){
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int num : nums) {
min = Math.min(min, num);
max = Math.max(max, num);
}
for (int i = 0; i < nums.length; i++) {
nums[i] -= min;
}
max -= min;
int maxLen = (max+"").length();
int[][] bucket = new int[nums.length][10];
int[] bucketCount = new int[10];
for (int i = 0, n = 1; i < maxLen; i++, n*=10) {
for (int num : nums) {
int digitVal = num / n % 10;
bucket[bucketCount[digitVal]][digitVal] = num;
bucketCount[digitVal]++;
}
int index = 0;
for (int j = 0; j < bucketCount.length; j++) {
if(bucketCount[j] > 0){
for (int k = 0; k < bucketCount[j]; k++) {
nums[index] = bucket[k][j];
index++;
}
}
bucketCount[j] = 0;
}
}
for (int i = 0; i < nums.length; i++) {
nums[i] += min;
}
}
11. 使用集合或 API
11.1 优先队列
public void priorityQueueSort(int[] nums){
PriorityQueue<Integer> queue = new PriorityQueue<>();
for(int num : nums){
queue.offer(num);
}
for (int i = 0; i < nums.length; i++) {
nums[i] = queue.poll();
}
}
11.2 Java API
public void arraysApiSort(int[] nums){
Arrays.sort(nums);
}