how broad first search works

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leetCode 45 (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.Your goal is to reach the last index in the minimum number of jumps.

e.g. [2, 3, 1, 1, 4] –> 2

solution: at current position, bfs all the next positions it can jump to, when the back element is in next position list, then reached.

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int sol(vector<int>& nums )
{
	vector<int>::iterator  it, cur_it ;
	vector<int> *adj = new vector<int>[nums.size()]; 	
	int cur_node ; 
	for(it=nums.begin(); it < nums.end();  it++)
	{
		cur_node = *it; 
		adj[cur_node].push_back(cur_node); 
		for(int i=1; i<=cur_node; i++)
		{
			cur_it =  std::next(it, i);
			adj[cur_node].push_back(*cur_it); 
			if( *cur_it == nums.back()){
				it = nums.end();  //reach the last element, so return 
			}
		}
	}
	
	vector<int> tmp; 
	int last_element = nums.back(); 
	int mini_path = 1; 
	for(int i=0; i < nums.size(); i++)
	{
		tmp = adj[i]; 
		if(!tmp.empty())
		{ 
			for(int j=0; j<tmp.size(); j++)
			{
				cout<< tmp[j] << ' ' ;
			}
			cout << endl; 
			if(tmp.back() == last_element)
			{
				return mini_path; 
			}
			mini_path++; 
		}
	}	
	return -1;
}