Jump Game

The problem can be think of as if you walk into i, how many available steps you have?
if you have 0 steps left while the current value is 0, then you won't be able to go to the next step.

The the question is how to calculate the current step, it is:
max(--nlen,nums[i])
since we reuse nlen, so I write it into two steps.


class Solution {
public:
    bool canJump(vector<int>& nums) {
        int n = nums.size();
        if ( n == 1 ) return true;

        int nlen = nums[0];
        
        for ( int i = 1; i<n; i++)
        {

            if (nlen <= 0 && nums[i-1] == 0 ) return false;
            nlen --;
            nlen = max(nlen,nums[i]);
        }

        return true;
            
    }
};