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...
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]. 题目意思:给出一个数组你需...
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 and 1). To the right of 2 there is on...
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...
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: Input: [5,2,6,1]Output: [2,1,1,0]Explanation:To the right of 5 there are 2 ...
【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...
[leetcode] 315. Count of Smaller Numbers After Self Description 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]....
elements()) #Counter的用法 from collections import Counter Counter('spam') #= Counter({'s': 1, 'p': 1, 'a': 1, 'm': 1}) #显示所有元素并记数 temp=Counter(ans) #Counter({'e': 1, 'l': 2}) list(temp.elements())=['e', 'l', 'l'] #显示所有元素 list一定要有 list(...
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...
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:...