Jump Game II

Problem:

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

    有点像接力赛，需要挑选潜力最大的下一棒。如果到达当前节点需要的跳数是最少的，如何选择下一跳呢？如果下一跳能够到达终点，那就直接跳到终点，如果不能到达终点，那就从当前节点覆盖的范围内选择跳得最远的那一个节点。

Solution:

public class Solution {
    public int jump(int[] A) {
     if(A==null||A.length<=1)
         return 0;

     int max_reach = A[0];
     int min_step = 1;
     int pos = 0;
     while(max_reach<A.length-1)
     {
         int max = max_reach;
         for(int i=pos+1;i<=max_reach;i++)
             if(i+A[i]>max)
             {
                 max = i+A[i];
                 pos = i;
             }
         max_reach = max;
         min_step++;
      }
      return min_step;
    }
}