Algorithm 通过对合并排序的递归调用使用控制流理解递归
我正在研究在合并排序中使用递归的控制流 我使用的特定算法是:Algorithm 通过对合并排序的递归调用使用控制流理解递归,algorithm,sorting,recursion,merge,Algorithm,Sorting,Recursion,Merge,我正在研究在合并排序中使用递归的控制流 我使用的特定算法是: MergeS(ar, p, r){ 1. if p<r{ 2. k = floor[(p+r)/2] 3. MergeS(ar, p , k) 4. MergeS(ar, k+1, r) //in the schematic diagram I have written this as mergeS(,,) 5. Merge(ar, p ,k, r) 6. } 7.} Merge(ar
MergeS(ar, p, r){
1. if p<r{
2. k = floor[(p+r)/2]
3. MergeS(ar, p , k)
4. MergeS(ar, k+1, r) //in the schematic diagram I have written this as mergeS(,,)
5. Merge(ar, p ,k, r)
6. }
7.}
Merge(ar, p, k, r){
8. n1 = k-p+1
9. n2 = r-p
10. let L[1...n1+1] and R[1....n2+1] be new arrays
11. for i=1 to n1
12. L[i] = ar[p+i-1]
13. for j=1 to n2
14. R[j] = ar[k+j]
15. L[n1+1] = Infinity
16. R[n2+1] = Infinity
17. i = 1
18. j = 1
19. for t = p to r
20. if L[i] <= R[j]
21. A[t] = L[i]
22. i = i+1
23. else
24. A[t] = R[j]
25. j = j+1
30. }
---暂停1:合并(ar,0,5)----调用-->2:合并(ar,0,2)---------
if(03:合并(ar,0,1)---------
if(04:合并(ar,0,0)---------
if(05:合并(ar,0+1,5)---------
if(16:合并(ar,1,3)---------
if(17:合并(ar,1,2)---------
if(18:合并(ar,1,1)---------
if(19:合并(ar,1+1,5)---------
if(210:合并(ar,2,3)---------
if(211:合并(ar,2,2)----------
if(212:合并(ar,2+1,5)----------
if(313:合并(ar,5,5)---------
if(5合并(ar,3,4,5)----------
n1=4-3+1=2
n2=5-4=1
对于(i=0到i=1):
迭代1:
L[0]=ar[3+0-1]=ar[2]//值56分配给ar[2]
迭代2:
L[1]=ar[3+1-1]=ar[3]//值12分配给ar[3]
对于(j=0到j=0):
迭代1:
R[0]=ar[4+0]=ar[4]//值4分配给ar[4]
x=0;
y=0;
对于(t=0到t=1):
迭代1:
/*如果(L[0]问题中的算法中的索引需要在代码中修改以便于使用。我不是说算法是错误的,它是绝对正确的,只需要修改
合并的调整代码
:
private static void Merge(int[] ar, int p, int k, int r) {
int n1 = k-p+1;
int n2 = r-k;
int[] L = new int[n1+1];
int[] R = new int[n2+1];
for(int i=0; i<n1; i++){
L[i] = ar[p+i]; // p+i and not p+i-1
}
for(int i=0; i<n2; i++){
R[i] = ar[k+i+1]; //k+1+i and not k+i
}
// Use of Tnteger.MAX_VALUE
L[n1]=Integer.MAX_VALUE;
R[n2]=Integer.MAX_VALUE;
int i = 0;
int j = 0;
for(int t=p; t<=r; t++){ //carefully set initialization and t<=r and not simply t<r
if(L[i]<r[j]){
ar[t] = L[i];
i++;
}
else{
ar[t] = R[j];
j++;
}
}
}
}
私有静态无效合并(int[]ar,int p,int k,int r){
int n1=k-p+1;
int n2=r-k;
int[]L=新的int[n1+1];
int[]R=新的int[n2+1];
对于(int i=0;显示的递归调用流显示从0开始的索引,但psuedo代码显示从1开始的索引。使用p=开始索引和r=结束索引(数组最后一个元素后的1)可能更简单。两个合并递归调用将是合并(ar,p,k)和合并(ar,k,r)…(k而不是k+1)。如果我采用合并(ar、k、r),则中间索引元素将不必要地被复制多次(当调用分区的右侧部分时)在每个递归调用中。如果您可以检查我在每个调用中传递的参数并验证它们是否正确?如果不正确,我在哪个调用中输入了错误的参数。假设r是合并中的结束索引(ar,p,r),那么整个数组的基于C的循环将是for(I=0;Iif(0<2):true ; set k=(0+2)/2
if(0<1):true ; set k=(0+1)/2
if(0<0):false
if(1<5):true ; set k=(1+5)/2
if(1<3):true ; set k=(1+3)/2
if(1<2):true ; set k=(1+2)/2
if(1<1):false
if(2<5):true ; set k=(2+5)/2
if(2<3):true ; set k=(2+3)/2
if(2<2):false
if(3<5):true ; set k=(3+5)/2
if(5<5):false
n1 = 4-3+1=2
n2 = 5-4=1
for(i = 0 to i=1):
iteration1:
L[0] = ar[3+0-1]= ar[2] //value 56 is assigned to ar[2]
iteration2:
L[1] = ar[3+1-1] = ar[3] //value 12 is assigned to ar[3]
for(j=0 to j=0):
iteration1:
R[0] = ar[4+0] = ar[4] //value 4 is assigned to ar[4]
x = 0;
y = 0;
for(t=0 to t=1):
iteration1:
/* if(L[0]<= R[0]) //56<=4:false */
else
ar[0] = R[0] //4 is assigned to ar[0]
y = y+1
iteration2:
if(L[0]<=R[1]): //56<=infinity
ar[1] = L[0] // 56 is assigned to ar[1]
private static void Merge(int[] ar, int p, int k, int r) {
int n1 = k-p+1;
int n2 = r-k;
int[] L = new int[n1+1];
int[] R = new int[n2+1];
for(int i=0; i<n1; i++){
L[i] = ar[p+i]; // p+i and not p+i-1
}
for(int i=0; i<n2; i++){
R[i] = ar[k+i+1]; //k+1+i and not k+i
}
// Use of Tnteger.MAX_VALUE
L[n1]=Integer.MAX_VALUE;
R[n2]=Integer.MAX_VALUE;
int i = 0;
int j = 0;
for(int t=p; t<=r; t++){ //carefully set initialization and t<=r and not simply t<r
if(L[i]<r[j]){
ar[t] = L[i];
i++;
}
else{
ar[t] = R[j];
j++;
}
}
}
}
if(0<5):true ; set k=(0+5)/2
if(0<2):true ; set k=(0+2)/2
if(0<1):true ; set k=(0+1)/2
if(0<0):false
if(1<_1_):_false_ ;