Permutation Sequence

最新推荐文章于 2018-12-26 13:42:41 发布

zephyr_be_brave

最新推荐文章于 2018-12-26 13:42:41 发布

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分类专栏： leetcode习题分析文章标签：全排列算法第k个

本文链接：https://blog.youkuaiyun.com/zephyr_be_brave/article/details/8727195

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本文介绍了一种求解给定整数n和k时找到第k个排列序列的算法。通过列举和标记所有可能的排列，实现了从1到n的所有唯一排列。文章提供了详细的C++实现代码，并解释了如何使用辅助数组来简化问题。

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Permutation Sequence Mar 28 '12

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.

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

解法：算法见http://blog.youkuaiyun.com/zephyr_be_brave/article/details/9730967.实现时候注意用arr和index两个数组的实现方法。

class Solution {
public:
    string getPermutation(int n, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n == 0) return "";
        if(n == 1) return "1";
        k -= 1;
        vector<int> index(n,0);
        int factor = 1;
        for(int i = 1;i < n;i++)
        {
			factor *= i;
            index[i] = factor;  
        }
        
        
        for(int i = index.size()-1;i > 0;i--)
        {
            int temp = k/index[i];
            k -= temp * index[i];
            index[i] = temp;  
        }
        
        string res;
        vector<int> arr;
        for(int i = 0; i< n;i++) arr.push_back(i+1);
        for(int i = index.size()-1;i >= 0;i--)
        {
            res.append(1,'0'+ arr[index[i]]);
            arr.erase(arr.begin()+index[i]);
        }
   
        return res;
        
    }
};

20 milli secs.