ACM simulation questions (4) -- Recursion

Recursion is very intuitive to solve some problems, but you should pay attention to the depth of recursion when using recursion. If the depth is too deep, Stack Overflow may occur. The following example shows how to use it.

Question: Super stair
Problem Description
There is a total of M-level stairs. At the beginning, you are at the first level. If you can only cross the upper or second level at a time, how many steps should you go to the level M?
Input
The input data first contains an integer N, indicating the number of test instances, and then N rows of data. Each row contains an integer M (1 <= M <= 40), indicating the level of the stairs.
Output
For each test instance, please output the number of different steps
Sample Input
2
2
3
Sample Output
1
2
 
Solution: first, we need to find the rules for these questions. We can try to write a few M cases, and then find the rules from them. For example:
M walk
1
2 1-2
3 1-3 (two levels from 1) 1-2-3 (one level from 2)
4 1-2-4 (two levels from 2) 1-2-3-4 1-3-4 (one level from 3)
5 1-2-3-5 1-3-5 (two steps from 3) 1-2-3-4-5 1-2-4-5 1-3-4-5 (level from 4)
6 1-2-4-6 1-2-3-4-6 1-3-4-6 (two steps from 4)
1-2-3-5-6 1-3-5-6 1-2-3-4-5-6 1-2-4-5-6 1-3- 4-5-6 (from 5 to 5)
...
All the steps of n N-2 layer can reach n layer in two levels, and all the steps of n-1 layer can reach n layer
N layer walk can be expressed as f (n), then f (n) = f (n-1) + f (n-2), n> 4
In this way, Recursive Computing can be used. The reference code is as follows:
/*
* Super stair
*/
Public static int test7 (int n ){
Switch (n ){
Case 1: return 0;
Case 2: return 1;
Case 3: return 2;
Default: return test7 (n-1) + test7 (n-2 );
}
}
For such a question, recursion is not fast, and it will overflow if recursion is too deep. You can use a loop to process it. refer to the following code:
/*
* Super stair
*/
Public static int test7 (int n ){
Int [] result = new int [40];
Result [0] = 0;
Result [1] = 1;
Result [2] = 2;
For (int I = 3; I <n; I ++ ){
Result [I] = result [I-1] + result [I-2];
}
Return result [n-1];
}
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