【Leetcode】Permutation Sequence

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题目链接：https://leetcode.com/problems/permutation-sequence/

题目：

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.

思路：

从左往右计算，先计算最高位。 应该为k/(n-1)! ，下一位应该是(k%(n-1)!)/(n-2)!，以此类推。 

算法：

public String getPermutation(int n, int k) {  
        int nums[] = new int[n];  
        int pCount=1;  
        for(int i=0;i<nums.length;i++){  
            nums[i] = i+1;  
            pCount *= (i+1);//n!  
        }  
        k--;//下标从0开始  
        String res = "";  
        for(int i=0;i<n;i++){  
            pCount /=n-i;//(n-i-1)!  
              
            int r = k/pCount;  
            res+=nums[r];//第r小的元素  
            for(int j=r;j<n-1-i;j++){ //获取第r小的元素 所以将已经用过的元素覆盖    
                nums[j] = nums[j+1];  
            }  
            k = k%pCount;  
        }  
        return res;  
    }  

【Leetcode】Permutation Sequence

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原文地址：http://blog.csdn.net/yeqiuzs/article/details/51542413