Count of Smaller Numbers After Self

最新推荐文章于 2025-12-03 23:35:37 发布

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本文介绍了一种高效算法，解决计数数组中每个元素右侧更小元素的问题。通过逆序遍历并使用带有计数功能的二叉查找树，实现了时间复杂度远低于简单搜索的方法。

首先吐槽下优快云，写完，发表，直接没了，辛苦写的东西就这么不见了……

You are given an integer array nums and you have to return a new counts array. Thecounts 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].

这道题目的意思就是输入数组nums，输出数组arr，其中arr[i]的值是数组nums中位于nums[i]右侧且小于nums[i]的元素是数目，如果暴力搜索的话，时间复杂度大概为O(n^2)，太高了。题目是比较位于右侧的元素，因此我们逆序遍历数组nums，并把大小信息用二叉查找树存储起来。在构造二叉查找树的时候，为每个二叉查找树节点增加一个属性用于记录子树节点数目，便于后续统计符合要求的元素数目。同时，需要考虑数组中值相同的元素，由于统计的是小于目标的元素数目，因此把相等的值也插入节点的右子树中。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;
	}