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):
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.
» Solve this problem[Thoughts]
两个解法。
第一,DFS
递归遍历所有可能,然后累加计算,直至到K为止。
第二,数学解法。
假设有n个元素,第K个permutation是
a1, a2, a3, ..... ..., an
那么a1是哪一个数字呢?
那么这里,我们把a1去掉,那么剩下的permutation为
a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列,那么这里就可以知道
设变量K1 = K
a1 = K1 / (n-1)!
同理,a2的值可以推导为
a2 = K2 / (n-2)!
K2 = K1 % (n-1)!
.......
a(n-1) = K(n-1) / 1!
K(n-1) = K(n-2) /2!
an = K(n-1)
实现如下:
1: string getPermutation(int n, int k) {
2: vector<int> nums(n);
3: int permCount =1;
4: for(int i =0; i< n; i++)
5: {
6: nums[i] = i+1;
7: permCount *= (i+1);
8: }
9: // change K from (1,n) to (0, n-1) to accord to index
10: k--;
11: string targetNum;
12: for(int i =0; i< n; i++)
13: {
14: permCount = permCount/ (n-i);
15: int choosed = k / permCount;
16: targetNum.push_back(nums[choosed] + '0');
17: //restruct nums since one num has been picked
18: for(int j =choosed; j< n-i; j++)
19: {
20: nums[j]=nums[j+1];
21: }
22: k = k%permCount;
23: }
24: return targetNum;
25: }
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