1409. Queries on a Permutation With Key

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

• In the beginning, you have the permutation P=[1,2,3,...,m].
• For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].

Return an array containing the result for the given queries.

Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1] 
Explanation: The queries are processed as follow: 
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. 
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. 
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. 
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. 
Therefore, the array containing the result is [2,1,2,1].  

Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]

Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]

Constraints:

• 1 <= m <= 10^3
• 1 <= queries.length <= m
• 1 <= queries[i] <= m

class Solution {
    public int[] processQueries(int[] queries, int m) {
        List<Integer> list1 = new ArrayList();
        int le = queries.length;
        for(int i = 1; i <= m; i++) list1.add(i);
        
        List<Integer> res = new ArrayList();
        for(int i = 0; i < le; i++){
            res.add(list1.indexOf((Integer) queries[i]));
            list1.remove((Integer) queries[i]);
            list1.add(0, (Integer) queries[i]);
        }
        int[] result = new int[le];
        for(int i = 0; i < le; i++) result[i] = res.get(i);
        return result;
    }
}

注意，list.remove() 有两种，一种是remove(index )， 另一个是remove(Object o), 此处应该用Object o