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.

貪心法，一直往前走，當找到比之前能走的更遠的位置max_pos 則更新即可。最後判斷max_pos是否能覆蓋n-1

 bool jump_game(int *a,int n)
    {
    	if (a == NULL || n <= 0)
    	{
    		return false;
    	}
    
    	int max_pos = 0;
    
    	for (int i = 0;i <= max_pos && i < n;i++)
    	{
    		if (max_pos < a[i] + i)
    		{
    			max_pos = a[i] + i;
    		}
    	}
    	//cout<<max_pos<<endl;
    	return max_pos >= n-1;
    }
    

第二次;

跟上面類似，貪心，只是此處用的是步數，略微麻煩。一直往前走，每次更新當前的步數（1、走一步減去一步。2、比之前大跟新.）（下一步開始）

bool jump_game_2(int *a,int n)
    {
        	if (a == NULL || n <= 0)
    	{
    		return false;
    	}
    
    	int max_step = a[0];//以當前最大步數決定！
    	
    	for (int i = 1; i < n ;i++)
    	{
    		if (max_step > 0)//下一個開始減
    		{
    			max_step -- ;
    			if (max_step < a[i])
    			{
    				max_step = a[i] ;
    			}
    		}
    		else
    		{
    			return false;
    		}	
    	}
	 
	    return true;
    }

貪心法。