1409.Queries on a Permutation With Key

Tags: Medium Array

Links: https://leetcode.com/problems/queries-on-a-permutation-with-key/

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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

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class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        vector<int> num(m);
        for (int i = 0; i < m; ++i) num[i] = i + 1;
        int n = queries.size();
        vector<int> res(n);

        for (int i = 0; i < n; ++i) {
            int target = queries[i];
            int pos = find(num, target);
            res[i] = pos;
        }

        return res;
    }

    int find(vector<int> & num, int target)
    {
        int n = num.size();
        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (num[i] == target) {
                res = i; break;
            }
        }

        int pos = res;
        int tmp = num[pos];
        for (int i = pos; i >= 1; --i) {
            num[i] = num[i - 1];
        }
        num[0] = tmp;

        return res;
    }
};

暴力模拟。