Sorting algorithms are so common and so basic that in most cases they are integrated into the language's auxiliary libraries. Although the sorting algorithm can be used very conveniently, understanding the sorting algorithm can help us find the direction of solving the problem.
1. Bubble Sort
Bubble sort is one of the simplest and crudest sorting methods. Its principle is very simple. Each time the pair is compared from left to right, the larger one is swapped to the back, and each time it can ensure that the maximum value of the first M elements is moved to the far right.
step
- Compare two adjacent elements x and y from the left, and swap the two if x > y
- perform a comparison and swap until the last element of the array is reached
- Repeat 1 and 2 until it is executed n times, that is, the n largest elements are all at the end
void bubble_sort(vector<int> &nums)
{
for (int i = 0; i < nums.size() - 1; i++) { // times
for (int j = 0; j < nums.size() - i - 1; j++) { // position
if (nums[j] > nums[j + 1]) {
int temp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = temp;
}
}
}
}
The step of the exchange can be done without temp, the method is
nums[j] += nums[j + 1];
nums[j + 1] = num[j] - nums[j + 1];
nums[j] -= num[j + 1];
Complexity Analysis
Since we have to repeatedly perform n times of bubbling, each bubbling has to perform n comparisons (actually an arithmetic sequence from 1 to n, that is (a1 + an) * n / 2), that is . The space complexity is . O(n^2)O(n)
2. Insertion Sort
The principle of insertion sort is to compare the selected number with the previous number from left to right, and find the most suitable position to put it in, so that the previous part is ordered.
step
- Starting from the left, select the number x at the current position, compare it with the number y before it, and swap the two if x < y
- Perform 1 step for all numbers before x until the previous numbers are in order
- Select a number after the ordered part and insert it into the previous ordered part until there are no numbers to choose from
void insert_sort(vector<int> &nums)
{
for (int i = 1; i < nums.size(); i++) { // position
for (int j = i; j > 0; j--) {
if (nums[j] < nums[j - 1]) {
int temp = nums[j];
nums[j] = nums[j - 1];
nums[j - 1] = temp;
}
}
}
}
Complexity Analysis
Because it has to select n times, and the worst is to compare n times when inserting, so the time complexity is the same O(n^2). The space complexity is O(n).
3. Selection Sort
The principle of selection sort is to find the largest (minimum) value from the out-of-order array each time, put it in the head of the current out-of-order array, and finally make the array in order.
step
- Starting from the left, select the minimum value of the following elements, and swap with the leftmost element
- Execute backwards from the current swapped position until the last element
void selection_sort(vector<int> &nums)
{
for (int i = 0; i < nums.size(); i++) { // position
int min = i;
for (int j = i + 1; j < nums.size(); j++) {
if (nums[j] < nums[min]) {
min = j;
}
}
int temp = nums[i];
nums[i] = nums[min];
nums[min] = temp;
}
}
Complexity Analysis
Each time, the minimum value is found once, and in the worst case, n times are found. Such a process needs to be performed n times, so the time complexity is still the same O(n^2). The space complexity is O(n).
4. Shell Sort
希尔排序从名字上看不出来特点，因为它是以发明者命名的。它的另一个名字是“递减增量排序算法“。这个算法可以看作是插入排序的优化版，因为插入排序需要一位一位比较，然后放置到正确位置。为了提升比较的跨度，希尔排序将数组按照一定步长分成几个子数组进行排序，通过逐渐减短步长来完成最终排序。
例子
例如 [10, 80, 70, 100, 90, 30, 20] 如果我们按照一次减一半的步长来算， 这个数组第一次排序时以3为步长，子数组是：
10 80 70 90 30 20 100
这里其实按照列划分的4个子数组，排序后结果为
10 30 20 90 80 70 100
也就是 [10, 30 20 90 80 70 100]
然后再以1为步长生成子数组
10 30 20 ..
这个时候就是一纵列了，也就是说最后一定是以一个数组来排序的。
步骤
- 计算当前步长，按步长划分子数组
- 子数组内插入排序
- 步长除以2后继续12两步，直到步长最后变成1
void shell_sort(vector<int> &nums)
{
for (int gap = nums.size() >> 1; gap > 0; gap >>= 1) { // times
for (int i = gap; i < nums.size(); i++) { // position
int temp = nums[i];
int j = i - gap;
for (; j >= 0 && nums[j] > temp; j -= gap) {
nums[j + gap] = nums[j];
}
nums[j + gap] = temp;
}
}
}
复杂度分析
希尔排序的时间复杂度受步长的影响，具体分析在维基百科。
5. 归并排序（Merge Sort）
归并排序是采用分治法（Divide and Conquer）的一个典型例子。这个排序的特点是把一个数组打散成小数组，然后再把小数组拼凑再排序，直到最终数组有序。
步骤
- 把当前数组分化成n个单位为1的子数组，然后两两比较合并成单位为2的n/2个子数组
- 继续进行这个过程，按照2的倍数进行子数组的比较合并，直到最终数组有序
void merge_array(vector<int> &nums, int b, int m, int e, vector<int> &temp)
{
int lb = b, rb = m, tb = b;
while (lb != m && rb != e)
if (nums[lb] < nums[rb])
temp[tb++] = nums[lb++];
else
temp[tb++] = nums[rb++];
while (lb < m)
temp[tb++] = nums[lb++];
while (rb < e)
temp[tb++] = nums[rb++];
for (int i = b;i < e; i++)
nums[i] = temp[i];
}
void merge_sort(vector<int> &nums, int b, int e, vector<int> &temp)
{
int m = (b + e) / 2;
if (m != b) {
merge_sort(nums, b, m, temp);
merge_sort(nums, m, e, temp);
merge_array(nums, b, m, e, temp);
}
}
这个实现中加了一个temp，是和原数组一样大的一个空间，用来临时存放排序后的子数组的。
复杂度分析
在merge_array过程中，实际的操作是当前两个子数组的长度，即2m。又因为打散数组是二分的，最终循环执行数是logn。所以这个算法最终时间复杂度是O(nlogn)，空间复杂度是O(n)。
6. 快速排序（Quick Sort）
快速排序也是利用分治法实现的一个排序算法。快速排序和归并排序不同，它不是一半一半的分子数组，而是选择一个基准数，把比这个数小的挪到左边，把比这个数大的移到右边。然后不断对左右两部分也执行相同步骤，直到整个数组有序。
步骤
- 用一个基准数将数组分成两个子数组
- 将大于基准数的移到右边，小于的移到左边
- 递归的对子数组重复执行1，2，直到整个数组有序
void quick_sort(vector<int> &nums, int b, int e, vector<int> &temp)
{
int m = (b + e) / 2;
if (m != b) {
int lb = b, rb = e - 1;
for (int i = b; i < e; i++) {
if (i == m)
continue;
if (nums[i] < nums[m])
temp[lb++] = nums[i];
else
temp[rb--] = nums[i];
}
temp[lb] = nums[m];
for (int i = b; i < e; i++)
nums[i] = temp[i];
quick_sort(nums, b, lb, temp);
quick_sort(nums, lb + 1, e, temp);
}
}
解法2: 不需要辅助空间
void quick_sort(vector<int> &nums, int b, int e)
{
if (b < e - 1) {
int lb = b, rb = e - 1;
while (lb < rb) {
while (nums[rb] >= nums[b] && lb < rb)
rb--;
while (nums[lb] <= nums[b] && lb < rb)
lb++;
swap(nums[lb], nums[rb]);
}
swap(nums[b], nums[lb]);
quick_sort(nums, b, lb);
quick_sort(nums, lb + 1, e);
}
}
复杂度分析
快速排序也是一个不稳定排序，时间复杂度看维基百科。空间复杂度是O(n)。
7. 堆排序（Heap Sort）
堆排序经常用于求一个数组中最大k个元素时。因为堆实际上是一个完全二叉树，所以用它可以用一维数组来表示。因为最大堆的第一位总为当前堆中最大值，所以每次将最大值移除后，调整堆即可获得下一个最大值，通过一遍一遍执行这个过程就可以得到前k大元素，或者使堆有序。
在了解算法之前，首先了解在一维数组中节点的下标：
- i节点的父节点 parent(i) = floor((i-1)/2)
- i节点的左子节点 left(i) = 2i + 1
- i节点的右子节点 right(i) = 2i + 2
步骤
- 构造最大堆（Build Max Heap）：首先将当前元素放入最大堆下一个位置，然后将此元素依次和它的父节点比较，如果大于父节点就和父节点交换，直到比较到根节点。重复执行到最后一个元素。
- 最大堆调整（Max Heapify）：调整最大堆即将根节点移除后重新整理堆。整理方法为将根节点和最后一个节点交换，然后把堆看做n-1长度，将当前根节点逐步移动到其应该在的位置。
- 堆排序（HeapSort）：重复执行2，直到所有根节点都已移除。
void heap_sort(vector<int> &nums)
{
int n = nums.size();
for (int i = n / 2 - 1; i >= 0; i--) { // build max heap
max_heapify(nums, i, nums.size() - 1);
}
for (int i = n - 1; i > 0; i--) { // heap sort
int temp = nums[i];
num[i] = nums[0];
num[0] = temp;
max_heapify(nums, 0, i);
}
}
void max_heapify(vector<int> &nums, int beg, int end)
{
int curr = beg;
int child = curr * 2 + 1;
while (child < end) {
if (child + 1 < end && nums[child] < nums[child + 1]) {
child++;
}
if (nums[curr] < nums[child]) {
int temp = nums[curr];
nums[curr] = nums[child];
num[child] = temp;
curr = child;
child = 2 * curr + 1;
} else {
break;
}
}
}
复杂度分析
堆执行一次调整需要O(logn)的时间，在排序过程中需要遍历所有元素执行堆调整，所以最终时间复杂度是O(nlogn)。空间复杂度是O(n)。