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.

 Notice

This problem have two method which is Greedy and Dynamic Programming.

The time complexity of Greedy method 
 
is O(n).

The time complexity of Dynamic Programming method is O(n^2).

We manually set the small data set to allow you pass the test in both ways.
 This is just to let you learn how to use this problem in dynamic programming ways. 
If you finish it in dynamic programming ways, you can try greedy method to make it accept again.

Have you met 
 
this question in a real interview? Yes
Example
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

Tags 

先想好狀態, 在從頭想想初始化和遍歷到當前是與前面的狀態方程怎麽得到, 或者從最後一個想想也行

public boolean canJump(int[] A) {
        // wirte your code here
        //state
        boolean[] can = new boolean[A.length];
        
        //initialize
        can[0] = true;
        
        //function
        
        for(int i = 1; i < A.length; i++) {
            for (int j = 0; j < i; j++) {
                if (can[j] && j + A[j] >= i) {
                    can[i] = true;
                    break;
                }
            }
        }
        
        return can[A.length - 1];
    }

Jump Game