合并两个有序的子序,要求空间复杂度为O(1) - legendmaner - 博客园


合并两个有序的子序,要求空间复杂度为O(1) - legendmaner - 博客园
数组al[0,mid-1]和al[mid,num-1]是各自有序的,对数组al[0,num-1]的两个子有序段进行merge,得到al[0,num-1]整体有序。要求空间复杂度为O(1)。注:al[i]元素是支持'<'运算符的。
/*
 2 数组a[begin, mid] 和 a[mid+1, end]是各自有序的,对两个子段进行Merge得到a[begin , end]的有序数组。 要求空间复杂度为O(1)
 3 方案:
 4 1、两个有序段位A和B,A在前,B紧接在A后面,找到A的第一个大于B[0]的数A[i], A[0...i-1]相当于merge后的有效段,在B中找到第一个大于A[i]的数B[j],
 5 对数组A[i...j-1]循环右移j-k位,使A[i...j-1]数组的前面部分有序
 6 2、如此循环merge
 7 3、循环右移通过先子段反转再整体反转的方式进行,复杂度是O(L), L是需要循环移动的子段的长度
 8 */
void Reverse(int *a , int begin , int end )   //反转
13 {
14     for(; begin < end; begin++ , end--)
15         swap(a[begin] , a[end]);
16 }
17 void RotateRight(int *a , int begin , int end , int k)    //循环右移
18 {
19     int len = end - begin + 1;  //数组的长度
20     k %= len;
21     Reverse(a , begin , end - k);
22     Reverse(a , end - k + 1 , end);
23     Reverse(a , begin , end);
24 }
25 
26 // 将有序数组a[begin...mid] 和 a[mid+1...end] 进行归并排序
27 void Merge(int *a , int begin , int end )
28 {
29     int i , j , k;
30     i = begin;
31     j = 1 + ((begin + end)>>1);    //位运算的优先级比较低,外面需要加一个括号,刚开始忘记添加括号,导致错了很多次
32     while(i <= end && j <= end && i<j)
33     {
34         while(i <= end && a[i] < a[j])
35             i++;
36         k = j;   //暂时保存指针j的位置
37         while(j <= end && a[j] < a[i])
38             j++;
39         if(j > k)
40             RotateRight(a , i , j-1 , j-k);   //数组a[i...j-1]循环右移j-k次
41         i += (j-k+1);  //第一个指针往后移动,因为循环右移后,数组a[i....i+j-k]是有序的
42     }
43 }
44 
45 void MergeSort(int *a , int begin , int end )
46 {
47     if(begin == end)
48         return ;
49     int mid = (begin + end)>>1;
50     MergeSort(a , begin , mid);   //递归地将a[begin...mid] 归并为有序的数组
51     MergeSort(a , mid+1 , end);   //递归地将a[mid+1...end] 归并为有序的数组
52     Merge(a , begin , end);       //将有序数组a[begin...mid] 和 a[mid+1...end] 进行归并排序
53 }
接着又网上搜索工作,找到一个block swapping.
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