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直接插入排序过程： 例： def insert_sort(lists):
count = len(lists)
for i in range(1, count):
tmp = lists[i]
j = i - 1
while j >= 0 and lists[j]>tmp:
lists[j+1] = lists[j]
j -= 1
lists[j+1] = tmp
return lists
void direct_insert_sort(int *ar, int count);
void direct_insert_sort(int *ar, int count){
int tmp;
int i;
int j;
for (i=1; i < count; i++){
tmp = ar[i];
j = i-1;
while(j>=0 && (ar[j] > tmp) ){
ar[j+1] = ar[j];
j--;
}
ar[j+1] = tmp;
}
}
希尔排序过程： 评价： def shell_sort(lists): count = len(lists) step = count/2 while step>0: for i in range(step, count, step): tmp = lists[i] j = i - step while j >= 0 and lists[j] > tmp: lists[j+step] = lists[j] j -= step lists[j+step] = tmp step/=2 return lists
void inner_direct_insert_sort(int *ar, int count, int step);
void shell_sort(int *ar, int count);
void shell_sort(int *ar, int count){
int step;
for (step=count/2; step > 0; step/=2)
inner_direct_insert_sort(ar, count, step);
}
// 调用插入排序，但是这里需要改变步长。
void inner_direct_insert_sort(int *ar, int count, int step){
int tmp;
int i;
int j;
for (i=step; i < count; i+=step){
tmp = ar[i];
j = i-step;
while(j>=0 && (ar[j] > tmp) ){
ar[j+step] = ar[j];
j-=step;
}
ar[j+step] = tmp;
}
}
冒泡排序：哈哈最简单了 1. 从头开始，依次和自己后面的元素进行比较，交换 时间复杂度也很高O（N^2） def bubble_sort(lists): count = len(lists) for i in range(0, count): for j in range(i+1, count) if lists[i] > lists[j]: lists[i], lists[j] = lists[j], lists[i] return lists
void bubble_sort(int *ar, int count);
void bubble_sort(int *ar, int count){
int i;
int j;
int tmp;
for(i=0; i<count; i++){
for (j=i+1; j<count; j++){
if(ar[i] > ar[j]){
tmp = ar[i];
ar[i] = ar[j];
ar[j] = tmp;
}
}
}
}
快速排序过程： 2、递归调用基本的步骤
最差情况：完全逆序、完全顺序 def quick_sort(lists, left, right): if left >= right: return lists tmp = lists[left] start = left end = right while left < right: while left < right and lists[right] > tmp: right -= 1 if left < right: lists[left] = lists[right] left += 1 while left < right and lists[left] < tmp: left += 1 if left < right: lists[right] = lists[left] right -= 1 lists[left] = tmp quick_sort(lists, start, left-1) quick_sort(lists, left+1, end) return lists
int base_action(int *ar, int start_index, int end_index);
void inner_quick_sort(int *ar, int start_index, int end_index);
void quick_sort(int *ar, int count);
void quick_sort(int *ar, int count){
inner_quick_sort(ar, 0, count-1);
}
void inner_quick_sort(int *ar, int start_index, int end_index){
int mid_index;
if(start_index < end_index){
mid_index = base_action(ar, start_index, end_index);
inner_quick_sort(ar, start_index, mid_index-1);
inner_quick_sort(ar, mid_index+1, end_index);
}
}
int base_action(int *ar, int start_index, int end_index){
int tmp;
tmp = ar[start_index];
while(start_index < end_index){
while(start_index < end_index && ar[end_index] > tmp){
end_index--;
}
if(start_index < end_index){
ar[start_index] = ar[end_index];
start_index++;
}
while(start_index < end_index && ar[start_index] < tmp){
start_index++;
}
if(start_index < end_index){
ar[end_index] = ar[start_index];
end_index--;
}
}
ar[start_index] = tmp;
return start_index;
}
直接选择排序过程： 评价： def select_sort(lists):
count = len(lists)
for i in range(count):
min_index = i
for j in range(i+1,count):
if lists[j] < lists[min_index]:
min_index = j
if min_index != i:
lists[i], lists[min_index] = lists[min_index], lists[i]
return lists
void direct_select_sort(int *ar, int count);
void direct_select_sort(int *ar, int count){
int tmp; //用于交换的中间值
int i; //下一个要比较的元素，即参考元素
int j; //除了已排好序的集合和下一个元素，剩下的所有元素的都和下一个元素比较
int minIndex; //最小的值的下标，将来放到已排好的元素中去
for (i=0; i < count-1; i++){
minIndex = i;
for (j = i+1; j<count; j++){
if(ar[j] < ar[minIndex] ){
minIndex = j;
}
}
if (minIndex != i){
tmp = ar[minIndex];
ar[minIndex] = ar[i];
ar[i] = tmp;
}
}
}
堆排序过程： 评价： def adjust_head(lists, root, count):
not_finished = True
while not_finished and root <= (count-1)/2:
max_index = root
left_child = 2*root + 1
right_child = 2*root + 2
if left_child < count and lists[left_child] > lists[max_index]:
max_index = left_child
if right_child < count and lists[right_child] > lists[max_index]:
max_index = right_child
if root != max_index:
lists[root], lists[max_index] = lists[max_index], lists[root]
else:
not_finished = False
root = max_index
def heap_sort(lists):
count = len(lists)
last_not_leaf_node = (count-1)/2
for root in range(last_not_leaf_node, -1, -1):
adjust_head(lists, root,count) #调整为大跟堆
while count > 1:
lists[count-1], lists[0] = lists[0], lists[count-1]
count -= 1
adjust_head(lists,root, count)
return lists
void adjustBigHeap(int *ar, int count, int root);
void heapSort(int *ar, int count);
void heapSort(int *ar, int count){
int root;
int tmp;
for(root = (count-1)/2-1; root > 0; root--){
adjustBigHeap(ar, count, root);
}
while(count > 1){
adjustBigHeap(ar, count, root);
tmp = ar[0];
ar[0] = ar[count-1];
ar[count-1] = tmp;
count--;
}
}
void adjustBigHeap(int *ar, int count, int root){
int maxIndex;
int tmp;
boolean finished = FALSE;
while(!finished && maxIndex < (count-1)/2){
maxIndex = 2*root+1 < count && ar[2*root+1] > ar[root] ? 2*root+1 : root;
maxIndex = 2*root+2 < count && ar[2*root+2] > ar[maxIndex] ? 2*root+2 : maxIndex;
if(maxIndex != root){
tmp = ar[root];
ar[root] = ar[maxIndex];
ar[maxIndex] = tmp;
}else{
finished = TRUE;
}
root = maxIndex;
}
}
归并算法：分而治之1. 将所有的数一分为二，在把其中的一组再一分为二，如此反复，最后在将有序的数组合并 2. 最关键的就是，递归终止的条件，即当每个组中只有一个元素。 3. 最重要的就是处理怎么合并。假设左边有序的起点为i, 右边有序的起点为j。将左右两边的数组相互比较，如果左边小的话，将比较的元素放入到结果集中，同时i应该加1。 如果右边元素大，则它放到结果集中，同时j要加1。如果左右两边其中的一边达到顶端(mid/end)，则把另一组元素全部放到结果集中 评价： 平均情况 最坏情况 最好情况 空间复杂度 def merge(left, right): i, j = 0, 0 rst = [] while i < len(left) and j < len(right): if left[i] <= right[j]: rst.append(left[i]) i += 1 else: rst.append(right[j]) j += 1 rst += left[i:] rst += right[j:] return rst def merge_sort(lists): if len(lists) <= 1: return lists middle = len(lists)/2 left = merge_sort(lists[middle:]) right = merge_sort(lists[:middle]) return merge(left, right)
void merge_sort(int *ar, int left, int right);
void merge(int *ar, int left, int right);
void merge_sort(int *ar, int left, int right){
int i;
if(left < right){
i = (left + right)/2;
merge_sort(a, left, i);
merge_sort(a, i+1, right);
merge(a, left, right);
}
}
void merge(int *ar, int left, int right){
int begin1 = left;
int mid = (left + right)/2;
int begin2 = mid+1;
int k = 0;
int new_ar_len = right - left + 1;
int *b = (int *)malloc(new_ar_len*sizeof(int));
while(begin1 <= mid && begin2 <= right){
if(ar[begin1] <= ar[begin2]){
b[k++] = ar[begin1++];
}else {
b[k++] = ar[begin2++];
}
while(begin1 <= mid)
b[k++] = ar[begin1++];
while(begin2 <= right)
b[k++] = ar[begin2++];
copy_array(b, a, new_ar_len, left);
free(b);
}
}
void copy_array(int *src, int *dst, int new_ar_len, int first){
int i;
int j=first;
for(i=0; i < len; i++){
dst[j] = src[i];
j++;
}
}
基数排序：将所有的数，按照各位进行排序并保持数组，接着按照十位排序，进行百位... 桶的个数，应该是基数的大小 进行比较的次数=所有元素最大数所占的位数 def radix_sort(lists):
for k in xrange(len(str(max(lists)))):
bucket = [ [] for _ in xrange(10)]
for i in lists:
bucket[i / (10 ** k) % 10].append(i)
lists = [element for item in bucket for element in item]
return lists
c语言我看着晕~
直接交换排序过程：
def exchange_sort(lists):
count = len(lists)
has_exchanged = True #在比较的过程中，如果没有数字发生交换，那么说明数据已经有序的了。退出即可
for i in range(0, count):
for j in range(0, count-i-1):
has_exchanged = False
if lists[j] > lists[j+1]:
lists[j], lists[j+1] = lists[j+1], lists[j]
has_exchanged = True
if not has_exchanged:
break
return lists
void direct_change_sort(int *ar, int count){
int i;
int j;
int tmp;
unsigned char has_exchanged;
for(i=0; i<count && has_exchanged ; i++){
for(j=0,has_exchanged = 0; j<count-i; j++){
if(ar[j] > ar[j+1]){
tmp = ar[j+1];
ar[j+1] = ar[j];
ar[j] = tmp;
has_exchanged = 1;
}
}
}
}
偷张图：
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2023-10-27
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