leetcode55—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.

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.

想法：动态规划方式，按照如下方式：

1、最后一步：假设在索引i的地方，能够到达索引n-1，那么需要满足条件(1)青蛙能够落在索引i(2)i+array[i]>n-1

2、状态转移表达式:f[j] = (0=<i<j)(f[i]&&i+f[i]>j)

3、边界条件：f[0]=true

4、计算顺序

class Solution {
public:
    bool canJump(vector<int>& nums) {
         == nums.size())
            return false;
        bool result[nums.size()] ;
        result[] = true;
//从第一块石头开始
         ; i < nums.size() ; i++){
            result[i] = false;
            //遍历石头i之前的所有石头，判断那一块能够落在石头i上
             ; j < i ; j++){
                if(result[j] && j + nums[j] >= i){
                    result[i] = true;
                }
            }
        }
        ];
    }

};