LeetCode 1803. Count Pairs With XOR in a Range (二叉樹)

題目

在一個數組裏面找到兩個數異或的結果在某個範圍之內。

這種題目，就要用二叉樹，

代碼寫的又臭又長。。

   struct Node
{
	int value;
	int left;
	int right;
	int num;
	int pos;
}tree[200005];

class Solution {
public:

int hight;
int lower;
int fun(int root, int number, int h, int l, int pos)
{
    if(pos==-1 )
        return tree[root].num;
	int b = number & (1 << pos);
    if(b>0) b=1;
	int ans = 0;
	
	int l1 = lower & (1 << pos);
	int h1 = hight & (1 << pos);
    if(l1>0) l1 =1;
    if(h1>0) h1 =1;
	if (h == 1 && l == 1)
	{
		ans = tree[root].num;
	}
	else if (h == 1 && l == 0)
	{
		if (b == 1 && l1 ==1)
		{            
			ans = fun(tree[root].left, number, h, l, pos - 1);
		}
		else if (b == 1 && l1 == 0)
		{
			ans = fun(tree[root].left, number, h, 1, pos - 1) + fun(tree[root].right, number, h, l, pos - 1);
		}
		else if (b == 0 && l1 == 1)
		{
			ans = fun(tree[root].right, number, h, l, pos - 1);
		}
		else if (b == 0 && l1 == 0)
		{
			ans = fun(tree[root].left, number, h, l, pos - 1) + fun(tree[root].right, number, h, 1, pos - 1);
		}
	}
	else if (h == 0 && l == 1)
	{
		if (b == 1 && h1 == 1)
		{
			ans = fun(tree[root].left, number, h, l, pos - 1) + fun(tree[root].right, number, 1, l, pos - 1);
		}
		else if (b == 1 && h1 == 0)
		{
			ans = fun(tree[root].right, number, h, l, pos - 1);
		}
		else if (b == 0 && h1 == 1)
		{
			ans = fun(tree[root].left, number, 1, l, pos - 1) + fun(tree[root].right, number, h, l, pos - 1);
		}
		else if (b == 0 && h1 == 0)
		{
			ans = fun(tree[root].left, number, h, l, pos - 1);
		}
	}
	else if (h == 0 && l == 0)
	{
		if (h1 == l1 && h1 == 0)
		{
			if (b == 1)
			{
				ans = fun(tree[root].right, number, h, l, pos - 1);
			}
			else if (b == 0)
			{
				ans = fun(tree[root].left, number, h, l, pos - 1);
			}
		}
		else if (h1 == l1 && h1 == 1)
		{
			if (b == 1)
			{
				ans = fun(tree[root].left, number, h, l, pos - 1);
			}
			else if (b == 0)
			{
				ans = fun(tree[root].right, number, h, l, pos - 1);
			}
		}
		else if (h1 == 1 && l1 == 0)
		{
          
			if (b == 1)
			{
				ans = fun(tree[root].left, number, h, 1, pos - 1) + fun(tree[root].right, number, 1, l, pos - 1);
			}
			else if (b == 0)
			{
				ans = fun(tree[root].left, number, 1, l, pos - 1) + fun(tree[root].right, number, h, 1, pos - 1);
			}
		}
    
	}


	return ans;
}
int pos2;
void init(int root, int p)
{
	if (p == 16)
		return;
	tree[root].left = ++pos2;
	tree[root].right = ++pos2;
    tree[root].num=0;
	init(tree[root].left, p + 1);
	init(tree[root].right, p + 1);
}
    
void init2(int root,int num, int p)
{
    tree[root].num++;
    if(p==-1)
        return;
    int b = num &(1<<p);
    if(b>0)
    {
        
        init2(tree[root].right,num,p-1);
    }
    else
    {
      
        init2(tree[root].left,num,p-1);
    }
}
int countPairs(vector<int>& nums, int low, int high) 
{
	pos2 = 0;
	init(0, 0);
	hight = high;
	lower = low;
	int res = 0;
	for (int i = 0; i < nums.size(); i++)
	{
		res += fun(0, nums[i], 0, 0, 14);
        init2(0,nums[i],14);
	}

	return res;
}
};