Leetcode Interview Question 62. The last remaining number in the circle

Problem Description

The n numbers 0,1, n-1 are arranged in a circle, starting with the number 0, and deleting the mth number from this circle each time. Find the last number left in this circle.

For example, the 5 digits 0, 1, 2, 3, and 4 form a circle, and the third digit is deleted every time from the digit 0, then the first 4 digits deleted are 2, 0, 4, 1, and so on. The remaining number is 3.

Input: n = 5, m = 3
Output: 3

Problem solving report

• The number of n person is $0,1,2,\cdots,n-1$ , delete the number as $m-1$ person, the next person deleted from $k+1$ starts counting, the corresponding number sequence is $m,m+1,\cdots,n-1,0,\cdots,m-2$。
• Remaining $n-1$Make a mapping of the number of$1$ person, and map it into $0,1,\cdots,n-2$ , then in the new sequence $x$ is the original sequence $(x+m) \% n$。
• in other words, $n-1$ person last remaining number is $x$ , then corresponding $n$The number of the last remaining person of$n$ people is $(x+m)\%n$。
• make $f(n)$ means $n$ number of the last person left, then $f(n)=(f(n)+m)\%n$。

Implementation code

• Recursive

class Solution {
	public:
		int f(int n, int m) {
	        if (n == 1)
	            return 0;
	        int x = f(n - 1, m);
	        return (m + x) % n;
	    }
    	int lastRemaining(int n, int m) {
        	return f(n, m);
    	}
};

• Iterate

class Solution {
public:
    int lastRemaining(int n, int m) {
        int ans=0;
        for(int i=2;i<=n;i++){
            ans=(ans+m)%i;
        }
        return ans;
    }
};