第一个成员开始正好走到数组最后一个成员所使用的最小步骤数（三种方法）

import java.util.*; public class Main {

static int min = Integer.MAX_VALUE;
public static void main(String[] args){
    Scanner in = new Scanner(System.in);
    while(in.hasNextInt()){
        int n = Integer.parseInt(in.next());
        int[] arr = new int[n];
        for(int i=0; i<n; i++){
            arr[i] = in.nextInt();
        }
        System.out.println(df(arr, 0, 0) +" " + dp(arr, 0) + " " + dt(arr));
    }
}

public static int df(int[] arr, int i, int k){
    if(arr.length <= 2) return 1;
    if(i == arr.length - 1) return k;
    if(i >= arr.length) return -1;
    if(i == 0){
        int min = arr.length;
        for(int j=1; j<arr.length/2; j++){
            int t = df(arr, i+j, k+1);
            if(t>0){
                min = Math.min(min, t);
            }
        }
        return min==arr.length?-1:min;
    }
    return df(arr, i+arr[i], k+1);
}

public static int dp(int[] arr, int i){
    if(arr.length <= 2) return 1;
    if(i == arr.length - 1) return 0;
    if(i >= arr.length) return -1;
    int res = arr.length;
    if(i == 0){
        for(int j=1; j<arr.length/2; j++){
            int t = dp(arr, i+j);
            if(t > 0){
                res = Math.min(res, t);
            }
        }
        res = res==arr.length?-1:res+1;
    } else {
        int t = dp(arr, i+arr[i]);
        res = t<0?-1:t+1;
    }
    return res;
}

public static int dt(int[] arr){
    if(arr.length <= 2) return 1;
    int[] dt = new int[arr.length];
    dt[0] = 0;
    dt[1] = 1;
    for(int i=2; i<arr.length; i++){
        if(i<arr.length/2){
            dt[i] = 1;
        } else{
            dt[i] = i+1;
            for(int j=1; j<i; j++){
                if(dt[j] < 0) continue;
                int count = 1;
                int index = j;
                while(index < i){
                    index += arr[index];
                    count++;
                }
                if(index == i){
                    dt[i] = Math.min(dt[i], count);
                }
            }
            if(dt[i] > i) dt[i] = -1;
        }
    }
    return dt[arr.length-1];
}

}