[leetcode] Jump Game 解题报告

题目链接：https://leetcode.com/problems/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.

动态规划的题目，保存一个当前最大可到达的距离maxreach，和一个当前这步可到达的距离比较,可以得到状态转移方程

max(maxreach, A[i] + i);

遍历一遍数组，如果到某一步的时候最大距离并不能到这里，则说明不能走到最后，时间复杂度为O(N), 空间复杂度为O(1), 代码如下：

class Solution {
private:
public:
    bool canJump(vector<int>& nums) {
        if(nums.size() ==0)
            return false;
        int maxreach = 0;
        for(int i =0; i< nums.size(); i++)
        {
            if(maxreach < i)
                return false;
            maxreach = maxreach > (nums[i] + i)?maxreach:(nums[i] + i);
        }
        return true;
    }
};