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1331. Rank Transform of an Array

作者:互联网

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:

 

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:

class Solution {
    public int[] arrayRankTransform(int[] arr) {
        Map<Integer, List<Integer>> map = new TreeMap();
        int rank = 1;
        
        for(int i = 0; i < arr.length; i++){
            map.putIfAbsent(arr[i], new ArrayList());
            map.get(arr[i]).add(i);
        }
        for(Map.Entry<Integer, List<Integer>> entry: map.entrySet()){
            List<Integer> cur = entry.getValue();
            for(int i: cur) arr[i] = rank;
            rank++;
        }
        return arr;
    }
}

虽然是个easy的题,但感觉还挺有意思

先用treemap把key从小到大排序,同时value用数组记录key的index

然后遍历每个key的value,然后把rank放入数组,rank++

标签:map,arr,int,1331,Transform,rank,element,Array,100
来源: https://www.cnblogs.com/wentiliangkaihua/p/12239575.html