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.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.



开始以为是dp, 其实不是.
用一个变量reach保存当前能够到达的最远距离索引.每次循环更新该索引即可. 额外需要注意的是循环的i不能超过这个reach. 因为这时无路可走了
class Solution {
public:
    bool canJump(vector<int>& nums) {
        int reach=0;
        for(int i=0;i<=reach&&i<nums.size();++i)
        {
            reach=max(reach,nums[i]+i);
        }
        return reach>=nums.size()-1;
    }
};