用c++实现java的tim排序 尽量还原
以下是用C++实现Java的Tim排序的代码:
#include <iostream>
#include <vector>
#include <algorithm>
const int RUN = 32;
void insertionSort(std::vector<int>& arr, int left, int right) {
for (int i = left + 1; i <= right; i++) {
int key = arr[i];
int j = i - 1;
while (j >= left && arr[j] > key) {
arr[j + 1] = arr[j];
j--;
}
arr[j + 1] = key;
}
}
void merge(std::vector<int>& arr, int left, int mid, int right) {
int len1 = mid - left + 1;
int len2 = right - mid;
std::vector<int> leftArr(len1);
std::vector<int> rightArr(len2);
for (int i = 0; i < len1; i++) {
leftArr[i] = arr[left + i];
}
for (int i = 0; i < len2; i++) {
rightArr[i] = arr[mid + 1 + i];
}
int i = 0;
int j = 0;
int k = left;
while (i < len1 && j < len2) {
if (leftArr[i] <= rightArr[j]) {
arr[k] = leftArr[i];
i++;
}
else {
arr[k] = rightArr[j];
j++;
}
k++;
}
while (i < len1) {
arr[k] = leftArr[i];
i++;
k++;
}
while (j < len2) {
arr[k] = rightArr[j];
j++;
k++;
}
}
void timSort(std::vector<int>& arr, int n) {
for (int i = 0; i < n; i += RUN) {
insertionSort(arr, i, std::min((i + RUN - 1), (n - 1)));
}
for (int size = RUN; size < n; size *= 2) {
for (int left = 0; left < n; left += 2 * size) {
int mid = left + size - 1;
int right = std::min((left + 2 * size - 1), (n - 1));
merge(arr, left, mid, right);
}
}
}
int main() {
std::vector<int> arr = {5, 2, 8, 9, 1, 3, 7, 4, 6};
int n = arr.size();
std::cout << "Original Array: ";
for (int i = 0; i < n; i++) {
std::cout << arr[i] << " ";
}
std::cout << std::endl;
timSort(arr, n);
std::cout << "Sorted Array: ";
for (int i = 0; i < n; i++) {
std::cout << arr[i] << " ";
}
return 0;
}
这段代码使用了插入排序(insertion sort)和归并排序(merge sort)来实现Tim排序(Timsort)。Tim排序是一种混合了插入排序和归并排序的稳定排序算法,它的特点是在处理小规模数据时具有较高的效率,同时也能处理大规模数据。在代码中,我们通过设置RUN的大小来确定何时使用插入排序。然后,我们使用归并排序将排好序的子数组合并为较大的子数组,直到整个数组排序完成
原文地址: https://www.cveoy.top/t/topic/iNJ8 著作权归作者所有。请勿转载和采集!