LeetCode(045) Jump Game II (Java)

题目如下：

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.)

分析如下：

隐藏着一个转化的结论。

如果想得到最少能够到达last index的步数，那么就等价于得到每步能跳跃的范围(min, max)，然后看看这个范围什么时候第一次到达last index。 这里也有一些分析。

我的代码：

// 
// Input:	[0]
// Output:	1
// Expected:	0
// 253 ms
public class Solution {
    public int jump(int[] A) {
        if (A.length <= 1) return 0; 
        if (A[0] == 0) return -1;
        int maxCoverIndex = A[0]; 
        int steps = 0, start = 0;
        for (; start < A.length && start <= maxCoverIndex; ) {
            ++steps;
            if (maxCoverIndex >= A.length - 1) {
                return steps;
            }
            int nextMax = 0;
            for (int i = start; i <= maxCoverIndex; ++i) {
                if ((i + A[i]) > nextMax) {
                    nextMax = i + A[i];
                    start = i;
                }
            }
            maxCoverIndex = nextMax;
        }
        return -1;
    }
}