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#yyds干货盘点# 解决名企真题：小招喵跑步
source link: https://blog.51cto.com/u_15488507/5488190
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#yyds干货盘点# 解决名企真题：小招喵跑步
原创1.描述：
描述小招喵喜欢在数轴上跑来跑去，假设它现在站在点n处，它只会3种走法，分别是：1.数轴上向前走一步，即n=n+1 2.数轴上向后走一步,即n=n-1 3.数轴上使劲跳跃到当前点的两倍，即n=2*n现在小招喵在原点，即n=0，它想去点x处，快帮小招喵算算最快的走法需要多少步？
输入描述：小招喵想去的位置x
输出描述：小招喵最少需要的步数
示例12.代码实现：
import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
System.out.println(solve(x));
}
public static int solve(int x){
if(x<2&&x>=0){
return x;
}
if(x < 0){
x = -x;
}
int[] dp=new int[x+1];
dp[0]=0;
dp[1]=1;
for(int i=2;i<=x;i++){
if(i%2==0){
dp[i]=dp[i/2]+1;
}else{
dp[i]=Math.min(dp[i-1], 1 + dp[(i + 1) / 2])+1;
}
}
return dp[x];
}
}
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
System.out.println(solve(x));
}
public static int solve(int x){
if(x<2&&x>=0){
return x;
}
if(x < 0){
x = -x;
}
int[] dp=new int[x+1];
dp[0]=0;
dp[1]=1;
for(int i=2;i<=x;i++){
if(i%2==0){
dp[i]=dp[i/2]+1;
}else{
dp[i]=Math.min(dp[i-1], 1 + dp[(i + 1) / 2])+1;
}
}
return dp[x];
}
}
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