归并排序

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1.经典

分治三部走（算法竞赛入门经典P225）:
划分问题、递归问题、合并问题；

#include<iostream>using namespace std;void merge_sort(int *A,int x,int y,int *T){//[x,y)左闭右开区间    if ((y-x)>1){//大于1，一个没必要划分再排序        int m = x + (y - x) / 2;        int p = x, q = m, i = x;        merge_sort(A,x,m,T);        merge_sort(A,m,y,T);        //int p = x, q = m, i = x;//写在这也可以        while (p<m||q<y){            if (q>=y||(p<m&&A[p]<=A[q])){                T[i++] = A[p++];            }            else{                T[i++] = A[q++];            }        }        for (i = x; i < y;i++){            A[i] = T[i];        }    }}int main(){    int A[] = {2,4,1,3,5};    int *T = new int[5];    for (int i = 0; i < 5; i++){        cout << A[i] << " ";    }    cout << endl;    merge_sort(A,0,5,T);    for (int i = 0; i < 5;i++){        cout << A[i] << " ";    }    cout << endl;    return 0;}

（两路）归并排序原理感觉比较难理解，先上两个图：
这里写图片描述

这里写图片描述

个人理解，就是把问题划分后再排序合并的过程。

时间复杂度为O(nlgn)；空间复杂度O(n);

动画演示：
http://www.ee.ryerson.ca/~courses/coe428/sorting/mergesort.html

另外一种比较容易记的实现方式

//将有二个有序数列a[first...mid]和a[mid...last]合并。void mergearray(int a[], int first, int mid, int last, int temp[]){    int i = first, j = mid + 1;    int m = mid,   n = last;    int k = 0;    while (i <= m && j <= n)    {        if (a[i] <= a[j])            temp[k++] = a[i++];        else            temp[k++] = a[j++];    }    while (i <= m)        temp[k++] = a[i++];    while (j <= n)        temp[k++] = a[j++];    for (i = 0; i < k; i++)        a[first + i] = temp[i];}void mergesort(int a[], int first, int last, int temp[]){    if (first < last)    {        int mid = (first + last) / 2;        mergesort(a, first, mid, temp);    //左边有序        mergesort(a, mid + 1, last, temp); //右边有序        mergearray(a, first, mid, last, temp); //再将二个有序数列合并    }}bool MergeSort(int a[], int n){    int *p = new int[n];    if (p == NULL)        return false;    mergesort(a, 0, n - 1, p);    delete[] p;    return true;}

2.原地归并

其他的都一样，关键是合并两个有序的数组的方法

#include <iostream>#include<string>#include<vector>using namespace std;void reverse(vector<int> &A, int left, int right){    int i = left, j = right;    while (i < j){        swap(A[i], A[j]);        i++;        j--;    }}void exchange(vector<int> &A, int i,int mid,int j){    reverse(A, i, mid - 1);    reverse(A, mid, j);    reverse(A, i, j);}void inPlaceMerge(vector<int>&A, int left, int mid, int last){    int i = left, j = mid, k = last;    while (i < j && j <= k){        while (i < j && A[i] <= A[j]){            i++;        }        int tmpmid = j;        while (j <= k && A[j] <= A[i]){            j++;        }        exchange(A, i, tmpmid, j-1);        i = tmpmid;//    }}int main(){    vector<int> A = { 0, 1, 5, 6, 9, 2, 3, 4, 7, 8 };    inPlaceMerge(A, 0, 5, A.size() - 1);    for (auto i : A){        cout << i << " ";    }    cout << endl;    return 0;}

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