这道 leetcode 的 hard 题,灵活的考察了对排序算法的理解. 1.题目分析 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
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: Count of Smaller Numbers After Self You are given an integer array nums and you have toreturnanewcounts 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...
Problem: You are given an integer arraynumsand you have to return a newcountsarray. Thecountsarray has the property wherecounts[i]is the number of smaller elements to the right ofnums[i]. Example: Input:[5,2,6,1]Output:[2,1,1,0] Explanation:To the right of 5 there are2smaller el...
【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...
The smaller numbers on the right of a number are exactly those that jump from its right to its left during a stable sort. So I do mergesort with added tracking of those right-to-left jumps. 1. 2. 自己错误的code class Solution(object): ...
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...
leetcode 315. 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 ...猜你喜欢LeetCode 315 - Count of Smaller Numbers After Self 原题...
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 计算后面较小数字的个数 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 = [...