error: I can pass the case in the Visual Studio 2010, the case is tough: [25000, ......., 2369], totally 1305 numbers..I can surely pass it and just one recursion(because I jump from the largest number, the first jump is 1304) but the leetcode tells me it's TLE, I really confused, could you help me? thank you~

// the index means the situation where the point is, and 0 -> A[index] means the steps we can jump,

we try the step from the largest to the 0.

```
class Solution {
public:
int flag;
bool canJump(int A[], int n) {
flag = 0;//indicator of found of way to jump game, when flag = 1
int tip[n];
memset(tip, 1, sizeof(tip));//used for pruning, the initial value is 1
if(n == 0)
return false;
else if(n == 1)
return true;
else
{
can(A, 0, n, tip);
if(flag != 0)
return true;
else
return false;
}
}
void can(int A[], int index, int n, int tip[])
{
int step;
int tmp;
if(flag != 0)//flag !0 means win the jump game, no need to continue to recursion.
return;
if(index == n - 1)// means can reach the last point.
{
flag++;
return;
}
else
{
if(flag != 0)//used for pruning
return;
if(A[index] == 0)//can not move any steps, just pruning
return;
if(tip[index] == 0)//means the previous try has confirmed that this point can not reach the termination, used for pruning
return;
if(A[index] + index >= n)//jump from the largest value, but the largest value maybe larger than the rest of steps, so here we make a pruning, let it not more than the rest of steps.
step = n - 1 - index;
else
step = A[index];
for(; step >= 1; step--)
{
tmp = index;//protect parameter "index", so here I used the copy one
tmp += step;
if(A[tmp] == 0 && tmp != n - 1)// if value is 0 and the tmp is not the last point, so just continue next
continue;
if(tmp < n)
{
can(A, tmp, n, tip);
if(flag != 0)// if after the above can(), the flag changed, no need to recursion next
break;
}
else
return;
}
if(flag == 0)//after all "tmp" validation, we can see whether the point tmp is valid. just for pruning
tip[tmp] = 0;
return;
}
return;
}
```

};