leetcode 1331. Rank Transform of an Array 详解 python3_def arrayranktransform(self, arr: list[int]) -> li-优快云博客

本文链接：https://blog.youkuaiyun.com/CSerwangjun/article/details/104094429

本文详细解析了如何使用哈希表解决数组元素排名转换的问题，通过排序和映射技巧，实现了高效的时间复杂度O(NlogN)。文章提供了两种Python实现方式，并附带完整代码示例。

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一.问题描述

Given an array of integers arr, replace each element with its rank.

The rank represents how large the element is. The rank has the following rules:

Rank is an integer starting from 1.
The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
Rank should be as small as possible.

Example 1:

Input: arr = [40,10,20,30]
Output: [4,1,2,3]
Explanation: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.

Example 2:

Input: arr = [100,100,100]
Output: [1,1,1]
Explanation: Same elements share the same rank.

Example 3:

Input: arr = [37,12,28,9,100,56,80,5,12]
Output: [5,3,4,2,8,6,7,1,3]

Constraints:

0 <= arr.length <= 105
-109 <= arr[i] <= 109

`二.解题思路`

主要用hash。

将arr排序，然后hash每个值对应的rank，

然后一次遍历原arr，通过rank映射得到rank值。

虽然简单，但是实现起来可以看出代码功底。

时间复杂度:O(NlogN) (主要是排序的复杂度)

更多leetcode算法题解法: 专栏 leetcode算法从零到结束或者 leetcode 解题目录 Python3 一步一步持续更新～

三.源码

#version 1
class Solution:
    def arrayRankTransform(self, arr: List[int]) -> List[int]:
        arr_s=sorted(list(set(arr)))
        rank={arr_s[i]:i+1 for i in range(len(arr_s))}
        return map(rank.get,arr)

#version 2
class Solution:
    def arrayRankTransform(self, arr: List[int]) -> List[int]:
        n=len(arr)
        if n==0:return []
        if n==1:return [1]
        rst=[]
        arr_s=sorted(arr)
        order_idx,t={},1
        for i in range(n):
            if arr_s[i] not in order_idx:
                order_idx[arr_s[i]]=t
                t+=1
        return [order_idx[arr[i]] for i in range(n)]