抓住那头牛:
农夫知道一头牛的位置，想要抓住它。农夫和牛都位于数轴上，农夫起始位于点N(0<=N<=100000)，牛位于点K(0<=K<=100000)。农夫有两种移动方式：
1、从X移动到X-1或X+1，每次移动花费一分钟
2、从X移动到2*X，每次移动花费一分钟
假设牛没有意识到农夫的行动，站在原地不动。农夫最少要花多少时间才能抓住牛？
Catch that cow：
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤N ≤ 100,000) on a number line and the cow is at a pointK (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any pointX to the pointsX- 1 orX + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
以上是题目，具体输入输出如下：
输入
两个整数，N和K。
输出
一个整数，农夫抓到牛所要花费的最小分钟数。
这道题看似简单，实则有坑。简单的递归是不行的，要优先广度搜索。
要点：
1. 避免重复计算已通过节点，需要一个数组保存visited[]
2. 步数累积较简单，维护一个数组step[]，子节点为父节点+1
3. 将子节点压入队列的之前就应该检查是否visited
4. 越界处理，不能为负数或者太大越界
两种方法，第一种更简单明了
package 序列找数;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
public class catch_cow {
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(solu2(5,17));
}
public static int solu(int n, int k){
if(n==k) return 0;
int max = 10000000;
//一个放接下来的数的队列
LinkedList<Integer> list = new LinkedList<Integer>();
int[] visited = new int[max];
int[] step = new int[max];
list.add(n);
step[n]=0;
int chid=0;
while(!list.isEmpty()){
int father = list.removeFirst();
for(int i =0;i<3;i++){
if(i==0) chid=father+1;
if(i==1) chid=father-1;
if(i==2) chid=father*2;
if(chid<0||chid>max) continue;
if(visited[chid]==0){
visited[chid]=1;
list.add(chid);
step[chid]=step[father]+1;
}
if(chid == k) return step[chid];
}
}
return -1;
}
public static int solu2(int n, int k){
if(n==k) return 0;
int max = 10000000;
//一个放接下来的数的队列
LinkedList<Integer> list = new LinkedList<Integer>();
int[] visited = new int[max];
int[] step = new int[max];
list.add(n);
step[n]=0;
visited[n]=1;
int chid=0;
while(!list.isEmpty()){
int father = list.removeFirst();
//step[father]=
if(father==k) return step[father];
if(father<0||father>max) continue;
//visited[father]=1;
if(visited[father+1]==0){
visited[father+1]=1;
step[father+1]=step[father]+1;
list.add(father+1);
}
if(father>0&&visited[father-1]==0){
visited[father-1]=1;
step[father-1]=step[father]+1;
list.add(father-1);
}
if(visited[father*2]==0){
visited[father*2]=1;
step[father*2]=step[father]+1;
list.add(father*2);
}
}return -1;
}
}