[leetcode] Jump Game II

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.)

首先想到的是用递归的方法去解，慢慢的遍历，但是显然效率比较低，最后的结果是大数据过不了！

class Solution {
public:
int jump(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int ministep=n-1;
int count=0;
int depth=0;
countstep(A,n,depth,count,ministep);
return ministep;
}
void countstep(const int A[],int n, int &depth, int& count, int&ministep){
if(A[depth]>=n-depth-1) {
if(ministep>count+1)
ministep=count+1;
return;
}
for(int i=1 ; i<A[depth]+1 ; i++){
count++;
depth=depth+i;
countstep(A , n , depth,count,ministep);
depth=depth-i;
count--;
}
}
};