315. Count of Smaller Numbers After Self Given an integer arraynums, returnan integer arraycountswherecounts[i]is the number of smaller elements to the right ofnums[i]. Example 1: Input:nums = [5,2,6,1]Output:[2,1,1,0]Explanation:To the right of 5 there are2smaller elements (2 a...
https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/138154/The-C%2B%2B-merge-sort-template-for-pairs-'i'-'j'-problem https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76611/Short-Java-Binary-Index-Tree-BEAT-97.33-With-Detailed-Explanation https:...
数组 counts 有该性质: counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。 示例1: 输入:nums = [5,2,6,1] 输出:[2,1,1,0] 解释: 5 的右侧有 2 个更小的元素 (2 和 1) 2 的右侧仅有 1 个更小的元素 (1) 6 的右侧有 1 个更小的元素 (1) 1 的右侧有 0 个更小的元素...
Leetcode 315. Count of Smaller Numbers After Self 解法一:Trie树 O((log32)*n ) 假设全部元素都是正数,把这些数按二进制位存进Trie数,每个结点都统计以这个点为根树的叶子的数量。 存数的时候从bit位高位开始存储。 假如一个数的一个比特位是1,那么与该比特位同一层的为0的结点的对应的叶子数就是部分...
来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/count-of-smaller-numbers-after-self 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 题意是给一个数组,请你返回一个等长的 list,表示当前位置的右侧小于 nums[i] 的元素个数。
packageleetcodeimport("sort""/halfrost/LeetCode-Go/template")// 解法一 线段树funccountSmaller(nums[]int)[]int{iflen(nums)==0{return[]int{}}st,minNum,numsMap,numsArray,res:=template.SegmentCountTree{},0,make(map[int]int,0),[]int{},make([]int,len(nums))fori:=0;i<len(nums);i...
的次数self.count = 1# 比当前元素值小的元素个数self.smaller = 0# 左子树self.left_child = None# 右子树self.right_child = Noneclass BinarySearchTree:def __init__(self):self.root = Nonedef insert(self, value):if self.root == None:self.root = Node(value)return 0else:return self._...
提交网址: https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 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: 代码语言:...
【Leetcode】Count of Smaller Numbers After Self 题目链接:https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 题目: counts[i]is the number of smaller elements to the right ofnums[i]. Example: Given nums=[5,2,6,1]To the right of5there are2smallerelements(2and1).To the...
Here similar problem on leetcode : https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/ → Reply Kr_Shubham 11 months ago, # | 0 You can also use Segment Tree, if array value is large in that case you have to normalize it down to <=n range. → Reply...