1, Title Description:
0,1,, n-1 which are arranged in a circle numbers n, starting from the number 0, delete the m-th digit each time from inside the circle. Find the circle where the last remaining digit.
For example, 0,1,2,3,4 five numbers in a circle, each deletion of three numbers starting from the number 0, the first four digits of the deleted sequence is 2,0,4,1, so the final the remaining number is 3.
2. Example:
Example 1:
Input: n = 5, m = 3
Output: 3
Example 2:
Input: n = 10, m = 17
Output: 2
restrictions:
. 1 <= n-<^ = 10. 5
. 1 <= m <= 10. 6 ^
3, solving ideas
(1) 0, ..., n-1 number is stored in the list set
(2) loops through the list the beginIndex (initial value 0) from the beginning to find the m-th element in the list of index value index
(3) deleting the m-th element list.remove (index)
(4) traversing the starting index list updated and recorded beginIndex variable
(5) repeating (2), (3), (4 ) step, list until only one element in
4, code implementation
Classes in java.util * Import;.
class Solution {
public int lastRemaining (n-int, int m) {
IF (n-<m. 1 || <. 1) {
return -1;
}
IF (n-==. 1) {
return 0;
}
// store the original number
the ArrayList <Integer> = new new List the ArrayList <> ();
for (int I = 0; I <n-; I ++) {
List.add (I);
}
int the beginIndex = 0;
the while (n->. 1 ) {
int size = list.size ();
// Get deleted numbered index
int index = (m% size the beginIndex + -. 1)% size;
index = index == -1 ? size -1 : index;
list.remove (index);
the index number of the start time //
beginIndex = index == size - 1 ? 0 : index;
--n;
}
return list.get(0);
}
}