题目：
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.)
题解：
参考了http://blog.csdn.net/linhuanmars/article/details/21356187，这道题和Jump Game都是利用动态规划的思想。区别是，上一道题维护的全局最优是maxcover，一旦maxcover大于总长度，那么说明能跳到结尾。
而这道题除了维护maxcover外，还需要考虑维护最小步数，最小步数的维护靠maxcover作为每一步能跳的长度，代码如下：
1
public
int jump(
int[] A) {
2 if(A== null||A.length==0)
3 return 0;
4
5 int maxcover = 0;
6 int step = 0;
7 int lastcover = 0;
8 for( int i = 0; i<=maxcover&&i<A.length;i++){
9 if(i>lastcover){
10 step++;
11 lastcover = maxcover;
12 }
13
14 if(A[i]+i>maxcover)
15 maxcover = A[i]+i;
16 }
17
18 if(maxcover<A.length-1)
19 return 0;
20 return step;
21 }
2 if(A== null||A.length==0)
3 return 0;
4
5 int maxcover = 0;
6 int step = 0;
7 int lastcover = 0;
8 for( int i = 0; i<=maxcover&&i<A.length;i++){
9 if(i>lastcover){
10 step++;
11 lastcover = maxcover;
12 }
13
14 if(A[i]+i>maxcover)
15 maxcover = A[i]+i;
16 }
17
18 if(maxcover<A.length-1)
19 return 0;
20 return step;
21 }