代码搜索：递归回溯

找到约 2,805 项符合「递归回溯」的源代码

代码结果 2,805
热门代码： main.c GPIO I2C SPI UART timer interrupt ADC PWM LED
www.eeworm.com/read/13112/268378

c 非递归.c

void main(); #include #define width (rings+1) void main() { int rings, last, next, x, z[500], s[3]; printf("how many rings? "); scanf("%d",&rings); for(x=1; x
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www.eeworm.com/read/13112/268416

c 阶乘递归.c

#include int factr(int n) { int result,r; if (n==1) return 1; result=factr(n-1)*n; return result; } main() { int a; a=factr(5); printf("%d",a); }
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www.eeworm.com/read/13112/268435

c 链表（递归）.c

#include #include struct listNode{ int data; struct listNode *nextPtr; }; typedef struct listNode LISTNODE; typedef LISTNODE * LISTNODEPTR; LISTNODEPTR list(LISTNODEPTR , int); /
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www.eeworm.com/read/13112/268442

c 递归车厢.c

/**********递归题改为非递归题实例 车厢********/ #include #define MAX 4 int stack[MAX],p=-1; struct { int num; int sign; }train[MAX]; void sub() { int inc; if(p==MAX-
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www.eeworm.com/read/14692/402486

txt 6.5递归.txt

STACK EQU 1FH BOTTOM EQU 00H M EQU 200 TOP DATA 3EH N DATA 30H NFACT EQU 31H TST: MOV SP,#5FH LCALL SETNULL MOV N,#5 LCALL FACT0
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www.eeworm.com/read/25789/949900

txt 6.5递归.txt

STACK EQU 1FH BOTTOM EQU 00H M EQU 200 TOP DATA 3EH N DATA 30H NFACT EQU 31H TST: MOV SP,#5FH LCALL SETNULL MOV N,#5 LCALL FACT0
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www.eeworm.com/read/491337/1191831

c 链表（递归）.c

#include #include struct listNode{ int data; struct listNode *nextPtr; }; typedef struct listNode LISTNODE; typedef LISTNODE * LISTNODEPTR; LISTNODEPTR list(LISTNODEPTR , int); /
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www.eeworm.com/read/458530/1585825

c 链表（递归）.c

#include #include struct listNode{ int data; struct listNode *nextPtr; }; typedef struct listNode LISTNODE; typedef LISTNODE * LISTNODEPTR; LISTNODEPTR list(LISTNODEPTR , int); /
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www.eeworm.com/read/181150/5278480

c 链表（递归）.c

#include #include struct listNode{ int data; struct listNode *nextPtr; }; typedef struct listNode LISTNODE; typedef LISTNODE * LISTNODEPTR; LISTNODEPTR list(LISTNODEPTR , int); /
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www.eeworm.com/read/327496/3455258

c 傻瓜递归.c

#include main() { int m=1,n=1,s; s=akm(m,n); printf("%d",s); } akm(int m,int n) { if(m==0) return n+1; else if(m!=0&&n==0) akm(m-1,1); else if(m!=0&&n!=0)
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