jump game

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.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

贪心策略，每一步局部最优 就可以 做到 全体最优，在每一次应用贪心策略时，都要验证贪心的正确性，这道题可以利用回溯法递归实现，但是会超时。因为状态空间树上遇到很多重复计算。

    bool canJump(vector<int>& nums) 
    {
        int len = nums.size();

        if(len <= 1)
        {
            return true;
        }

        int reach = 0;
        int pos;
        int i;
        int j;
        int M;

        for(i = 0; i < len;)
        {

            if(nums[reach] == 0)
            {
                return false;
            }
            j = 1;
            pos = reach;
            M = reach + 1 + nums[reach + 1];
            //cout << "*" << endl;

            while(j <= nums[reach])
            {

                if(nums[reach + j] + reach + j > M)
                {
                    if(nums[reach + j] + reach + j >= len - 1)
                    {
                        return true;
                    }
                    M = nums[reach + j] + reach + j;
                    pos = reach + j;
                }
                ++ j;
            }
            if(pos != reach)
            {
                reach = pos;
            }else
            {
                reach = pos + 1;
            }

            i = pos;
            //cout << pos << endl;
            if(nums[reach] == 0)
            {
                return false;
            }

            if(reach + nums[reach] >= len - 1)
            {
                return true;
            }

        }

    }

如果用递归，考虑下面的空间决策树