Count of Smaller Numbers After Self

最新推荐文章于 2025-07-18 19:55:42 发布

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最新推荐文章于 2025-07-18 19:55:42 发布

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分类专栏： LeetCode 文章标签：二叉树 leetcode

本文链接：https://blog.youkuaiyun.com/whucs2/article/details/51307308

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该博客主要介绍了如何利用二叉搜索树解决LeetCode的一道题目，即计算数组中每个元素右侧比其小的元素数量。通过从后向前遍历数组，将元素插入二叉搜索树的同时统计小于当前元素的数量，确保正确处理重复元素。

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You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Return the array

[2,
 1, 1, 0]

根据题意可以得知，对于输入数组nums，返回数组arrs，其中arrs[i]等于数组nums中位于nums[i]右边且小于nums[i]的元素数目。只需要保存与某个元素的比较结果，很容易想到二分查找树，所以，从后向前遍历数组nums，将元素nums[i]不断插入到二叉树中，同时统计当前小于nums[i]的元素数目。需要注意数组处理nums中的重复元素，由于统计小于nums[i]的元素，因此对于相等的元素也插入到二叉树节点右边，便于计算。代码如下

class BinarySearchTreeNode{
		BinarySearchTreeNode leftChildren;
		BinarySearchTreeNode rightChildren;
		int value;
		int num;
		public BinarySearchTreeNode(int v){
			this.value = v;
			num = 1;
		}
	}
	public int insert(int v, BinarySearchTreeNode node){
		int re = 0;
		while(true){
			if(v < node.value){
				node.num ++;
				if(node.leftChildren == null){
					node.leftChildren = new BinarySearchTreeNode(v);
					break;
				}
				node = node.leftChildren;
			}
			else{
				node.num ++;
				// v >= node.value,加上左子树以及node本身
				re += getNum(node.leftChildren);
				if(node.value != v)
					re++;
				if(node.rightChildren == null){
					node.rightChildren = new BinarySearchTreeNode(v);
					break;
				}
				node = node.rightChildren;
			}
		}
		
		return re;
	}
	
	private int getNum(BinarySearchTreeNode node){
		if(node == null)
			return 0;
		else
			return node.num;
	}
	
	public List<Integer> countSmaller(int[] nums) {
        LinkedList<Integer> reList = new LinkedList<Integer>();
        if(nums == null || nums.length == 0)
        	return reList;
        reList.addFirst(0);
        BinarySearchTreeNode root = new BinarySearchTreeNode(nums[nums.length-1]);
        for(int i=nums.length-2; i >= 0; i--){
        	int el = nums[i];
        	int re = insert(el,root);
        	reList.addFirst(re);
        }
        
        return reList;
	}