[leetcode]Permutation Sequence

Permutation Sequence

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):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

算法思路：

思路1：

最直观的思路，dfs逐个求，并计数。提交之后超时。

代码如下：

public class Solution {
    boolean success = false;
    StringBuilder str = new StringBuilder();
    int count = 0;
    public String getPermutation(int n, int k) {
        int count = 1;
        for(int i = 1; i <= n; count*=i++);
        if(k <= 0 || k > count ) return null;
        boolean[] hash = new boolean[10];
        for(int i = 1; i <= n;hash[i++] = true);
        StringBuilder sb = new StringBuilder();
        dfs(sb,hash,k,n);
        return str.toString();
    }
    private void dfs(StringBuilder sb,boolean[] set,int k,int n){
        if(sb.length() == n){
            count++;
            if(count == k) {
                success = true;
                str = sb;
            }
            return;
        }
        for(int i = 1; i <= n ; i++){
            if(!set[i]) continue;
            set[i] = false;
            sb.append(i);
            dfs(sb, set, k, n);
            if(success) return;
            set[i] = true;
            sb.deleteCharAt(sb.length() - 1);
        }
    }
    
}

思路2：

跳过中间那些无用的，直接求第k个。

寻找规律：

以n = 4,k = 20为例。以1为头的字符串一共有3!= 6个，同理以2、3开头的也有6个，因此第20个必定以4开头。

接下来只需要求以4开头的第2（20 - 18）个即可。

再递归处理。

public class Solution {
    public String getPermutation(int n, int k) {
        int count = 1;
        for(int i = 1; i <= n; count*=i++);
        if(k <= 0 || k > count ) return null;
        List<Integer> list = new ArrayList<Integer>();
        for(int i = 1; i <= n; list.add(i++));
        StringBuilder sb = new StringBuilder();
        helper(n, k, list, sb);
        return sb.toString();
    }
    
    public void helper(int n,int k ,List<Integer> list,StringBuilder sb){
        if(n == 1) {
            sb.append(list.get(0));
            return;
        }
        int count = 1;
        for(int i = 1; i <= n - 1;count *= i++);
        int index = 0;
        while(k > count){
            index++;
            k -= count;
        }
        sb.append(list.get(index));
        list.remove(index);
        helper(n - 1,k,list,sb);
    }    
}

思路3：

迭代处理

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