60. Permutation Sequence

最新推荐文章于 2020-06-01 17:59:09 发布

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最新推荐文章于 2020-06-01 17:59:09 发布

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分类专栏： leetcode

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本文介绍了一种解决LeetCode上第K个排列序列问题的高效算法，通过避免全排列的遍历，采用数学方法快速定位每个数位上的数字，从而实现O(N)的时间复杂度。

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60. Permutation Sequence

Total Accepted: 65886
Total Submissions: 249649
Difficulty: Medium

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the k^th permutation sequence.

刚开始用全排列的常规方法，DFS进行顺序查找，超时

public class Solution {
    static int index = 0;
    public String getPermutation(int n, int k) {
        int[] visited = new int[n];
        index = 0;
        return permutate(visited, n, k, "");
    }
    
    String permutate(int[] visited, int n, int k, String nowStr){
       
        String result = "";
        if(nowStr.length() == n){
            index++;
           return nowStr;
        }else{
            for(int i = 1; i <= n; i++){
                if(visited[i-1] == 0){
                    visited[i-1] = 1;
                    nowStr += i;
                    result = permutate(visited, n, k, nowStr);
                    //System.out.println(index);
                    if(index == k){
                        return result;
                    }
                    nowStr = nowStr.substring(0, nowStr.length()-1);
                    visited[i-1] = 0;
                }
            }
        }
        
        return result;
    }
    
}

转换思路，An, An-1, ..... , A2, A1 An能表示的数字有n！，利用此规律，第一次确定最高位，第二次确定次高位。。。。。。。循环即可得到结果

public class Solution {
    static int num = 0;
    public String getPermutation(int n, int k) {
        String result = "";
        int[] nums = new int[n+1];
        int[] visited = new int[n+1];
        nums[1] = 1;
        for(int i = 2; i <= n; i ++){
            nums[i] = i * nums[i-1];
        }
        
        for(int i = n; i >= 2; i --){
            int times = k / nums[i-1];
            if( k % nums[i-1] != 0){
                times++;
                k = k % nums[i-1];
            }else{
                k = nums[i-1];
            }
            int item = 1;
            for(int j = 1; j <= n; j++){
                if(visited[j] == 0){
                    if(times >= 0){
                        times--;
                    }
                    if(times == 0){
                        result += j;
                        visited[j] = 1;
                        break;
                    }
                }
            }
            
        }
        for(int j = 1; j <= n; j++){
            if(visited[j] == 0){
                result += j;
                visited[j] = 1;
            }
        }
        return result;
    }
  
    
}