Quicksort ( O(N*logN) )
It is more efficient in the same time complexity as ON*log(N).
The basic idea is:
1. First take a number from the sequence as the base number.
2. In the partitioning process, all the numbers larger than this number are placed on the right side of it, and all the numbers less than or equal to it are placed on the left side of it.
3. Then repeat the second step for the left and right intervals until each interval has only one number.
#include<bits/stdc++.h>
using namespace std;
//对于数组的传参，可以按照int arr[],也可以int *arr,这两种方法都是传入的地址，
//因此在函数中对数组的操作，会改变原有数组的值，
void quick_sort(int arr[],int begin,int end){
if ( begin<end) {
//挖走begin位置的数，这个位置的数会有别的数来填补
int i = begin, j = end, x = arr[begin];
//倒叙找到第一个小于x的数然后填入begin的坑
//这个时候j的位置就出现了一个新的坑，这个位置需要一个比基数大的数填
while (i < j && arr[j] >=x) j--;
if (i < j)arr[i++] = arr[j];
//从左到右找到第一个大于x的数，然后填补j位置的坑
while (i < j && arr[i] <= x)i++;
if (i < j)arr[j--] = arr[i];
arr[i] = x;
//分治，将i的左右两边再次快排
quick_sort(arr,begin,i-1);
quick_sort(arr,i+1,end);
}
}
int main() {
int arr[9] = {
1,1,34,5,7,8,9,0,11};
quick_sort(arr,0,8);
for (int i:arr) {
cout << i << endl;
}
return 0;
}
//结果：
//0
//1
//1
//5
//7
//8
//9
//11
//34
Selection sort (O(N^2))
Basic idea:
each time find a smallest number from the unsorted sequence and insert it into the last of the sorted sequence
void SelectSort(int arr[],int lenght) {
int index = 0;
while (index!=lenght) {
int x = 1e9, temp = 0;
for (int i = index; i <lenght;i++) {
if (arr[i]<=x) {
temp = i;
x = arr[i];
}
}
swap(arr[index],arr[temp]);
index++;
}
}
Merge sort (O(N*logN))
Divide and Conquer Ideas
Picture Demo
//归并排序
//分治，分到长度为2的最小单元，让这个单元有序，然后在从下往上开始有序
void Merge(vector<int> &arr ,int front,int mid,int end) {
//注意拷贝的区间。
vector<int> left(arr.begin()+front, arr.begin() + mid+1 );//拷贝左半部分
vector<int> right(arr.begin() + mid + 1,arr.begin()+end+1);//拷贝右半部分
//在尾部插入一个编译器允许的int类型最大值。
left.insert(left.end(),numeric_limits<int>::max());
right.insert(right.end(),numeric_limits<int>::max());
int leftindex = 0, rightindex = 0;
for (int i = front; i <= end;i++) {
if (left[leftindex] < right[rightindex])arr[i] = left[leftindex++];
else arr[i] = right[rightindex++];
}
}
//归并排序
void MergeSort(vector<int> &arr,int front,int end) {
if (front >= end) return;
int mid = (front + end) /2;
MergeSort(arr,front,mid);
MergeSort(arr,mid+1,end);
Merge(arr,front,mid,end);
}