leetcode 60. Permutation Sequence

最新推荐文章于 2020-07-19 10:42:57 发布

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本文介绍了一种求解第K个排列序列的有效算法。通过递减范围的方式，该算法能够在给定整数n和k的情况下，快速找到由1到n组成的唯一排列中的第k个序列。

60. 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):

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

Given n and k, return the kth permutation sequence.

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

思路在于缩小范围

class Solution {
public:
    string getPermutation(int n, int k) 
    {
        if (n == 1)
            return "1";
        vector<int> m1;
        string result;
        int p = 1;          //记录一共有多少种
        for(int i=1;i<=n;i++)
        {   
            p = p * i;
            m1.push_back(i);
        }
      
        for(int i = n; i > 0; i--)
        {      
            p = p / i;                              //以马上找到的这一位后面有多少种情况
            result = result + char(m1[(k-1)/p] + 48); 
            m1.erase(m1.begin() + (k - 1) / p);     //把插入过的删除
            k = k - (k-1)/p * p; 
        }
        return result;
    }
};