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· 3 min read

是一种混合稳定的排序算法,源自合并排序和插入排序,旨在较好地处理真实世界中各种各样的数据。

实现

// C++ program to perform TimSort.
#include<bits/stdc++.h>
using namespace std;
const int RUN = 32;

// This function sorts array from left index to
// to right index which is of size atmost RUN
void insertionSort(int arr[], int left, int right)
{
for (int i = left + 1; i <= right; i++)
{
int temp = arr[i];
int j = i - 1;
while (j >= left && arr[j] > temp)
{
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}

// Merge function merges the sorted runs
void merge(int arr[], int l, int m, int r)
{

// Original array is broken in two parts
// left and right array
int len1 = m - l + 1, len2 = r - m;
int left[len1], right[len2];
for (int i = 0; i < len1; i++)
left[i] = arr[l + i];
for (int i = 0; i < len2; i++)
right[i] = arr[m + 1 + i];

int i = 0;
int j = 0;
int k = l;

// After comparing, we
// merge those two array
// in larger sub array
while (i < len1 && j < len2)
{
if (left[i] <= right[j])
{
arr[k] = left[i];
i++;
}
else
{
arr[k] = right[j];
j++;
}
k++;
}

// Copy remaining elements of left, if any
while (i < len1)
{
arr[k] = left[i];
k++;
i++;
}

// Copy remaining element of right, if any
while (j < len2)
{
arr[k] = right[j];
k++;
j++;
}
}

// Iterative Timsort function to sort the
// array[0...n-1] (similar to merge sort)
void timSort(int arr[], int n)
{

// Sort individual subarrays of size RUN
for (int i = 0; i < n; i+=RUN) insertionSort(arr, i, min((i+RUN-1), (n-1)));

// Start merging from size RUN (or 32).
// It will merge
// to form size 64, then 128, 256
// and so on ....
for (int size = RUN; size < n; size = 2*size)
{

// pick starting point of
// left sub array. We
// are going to merge
// arr[left..left+size-1]
// and arr[left+size, left+2*size-1]
// After every merge, we
// increase left by 2*size
for (int left = 0; left < n;
left += 2*size)
{

// find ending point of
// left sub array
// mid+1 is starting point
// of right sub array
int mid = left + size - 1;
int right = min((left + 2*size - 1),
(n-1));

// merge sub array arr[left.....mid] &
// arr[mid+1....right]
if(mid < right)
merge(arr, left, mid, right);
}
}
}

// Utility function to print the Array
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
printf("\n");
}

// Driver program to test above function
int main()
{
int arr[] = {-2, 7, 15, -14, 0, 15, 0, 7, -7,
-4, -13, 5, 8, -14, 12};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Given Array is\n");
printArray(arr, n);

// Function Call
timSort(arr, n);

printf("After Sorting Array is\n");
printArray(arr, n);
return 0;
}

时间复杂度

O(n)=O(nlog2n)O(n)=O(n\log_2{n})

空间复杂度

O(n)O(n)

· 2 min read

http://datagenetics.com/blog/july22012/index.html

C++

#include <iostream>
#include <vector>
#include <string>
#include<algorithm>

using namespace std;

const int Floors = 25;
const int Eggs = 5;

// max{ fn{egg - 1, dropFloor} fn{egg, floor - dropFloor} }

// 一共egg个鸡蛋, floor层, 从dropFloor层扔, dropFloor 取值 1 - n
int drop(int egg, int floor) {
if (egg == 0 || floor == 0) {
printf("egg %d and floor %d: %d \n", egg, floor, 0);
return 0;
}

if (egg == 1) {
printf("egg %d and floor %d: %d \n", egg, floor, floor);
return floor;
}

if (floor == 1) {
printf("egg %d and floor %d: %d \n", egg, floor, 1);
return 1;
}

int minRes = INT_MAX;
for (int i = 1; i < floor; i++)
{
int curRes = max(drop(egg - 1, i - 1), drop(egg, floor - i));
if (minRes > curRes) {
minRes = curRes;
}
}

printf("egg %d and floor %d: %d \n", egg, floor, minRes + 1);
return minRes + 1;
}

int main()
{
int msg = drop(Eggs, Floors);

cout << msg << endl;
}

运行时间: 6.8512s

javascript

const Floors = 25;
const Eggs = 5;

// max{ fn{egg - 1, dropFloor} fn{egg, floor - dropFloor} }

// 一共egg个鸡蛋, floor层, 从dropFloor层扔
function drop(egg, floor) {
if (egg == 0 || floor == 0) {
return 0;
}

if (egg == 1) {
return floor;
}

if (floor == 1) {
return 1;
}

let minRes = Infinity;
for (let i = 1; i < floor; i++)
{
let curRes = Math.max(drop(egg - 1, i - 1), drop(egg, floor - i));
if (minRes > curRes) {
minRes = curRes;
}
}

return minRes + 1;
}

console.log(drop(Eggs, Floors));

运行时间: 6.022s

思路:F层E个鸡蛋 可以拆解成 假设从k层仍一个鸡蛋 Max{ k层E-1个鸡蛋, F-k层E个鸡蛋 }

k为最优解 可以用遍历的方式 k从 1 - F 层取值 Min{ (1 - F) 层扔一个鸡蛋 } 带入上面 递归计算。