Abstract : The Tower of Hanoi problem is the dividing line of IQ.
1. Code
Code first, then nonsense.
#include <stdio.h>
/**
* Hanoi.
*/
void hanoi(int paraN, char paraSource, char paraDestination, char paraTransit) {
if (paraN <= 0) {
return;
} else {
hanoi(paraN - 1, paraSource, paraTransit, paraDestination);
printf("%c -> %c \r\n", paraSource, paraDestination);
hanoi(paraN - 1, paraTransit, paraDestination, paraSource);
}// Of if
}// Of hanoi
/**
* Test the hanoi function.
*/
void hanoiTest() {
printf("---- addToTest begins. ----\r\n");
printf("2 plates\r\n");
hanoi(2, 'A', 'B', 'C');
printf("3 plates\r\n");
hanoi(3, 'A', 'B', 'C');
printf("---- addToTest ends. ----\r\n");
}// Of addToTest
/**
The entrance.
*/
void main() {
hanoiTest();
}// Of main
2. Running results
---- addToTest begins. ----
2 plates
A -> C
A -> B
C -> B
3 plates
A -> B
A -> C
B -> C
A -> B
C -> A
C -> B
A -> B
---- addToTest ends. ----
Press any key to continue
3. Code Description
- The recursive implementation of accumulation is purely for laying the foundation for this program.
- The Tower of Hanoi problem is very simple, and the function is fully calculated in only 9 lines; the Tower of Hanoi problem is very difficult, without figuring out a few parameters, it is directly dizzy in it.
- Here you can see the advantage of using meaningful, long variable names: program readability is greatly improved.
- The initial condition is 0 plates, so it is more convenient to judge the negative number of plates.
- The time complexity of the algorithm is O ( 2 n ) O(2^n)O(2n ), the space complexity isO ( n ) O(n)O ( n ) . All require drawing to explain.
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