60 Permutation Sequence

README.md

@@ -7,6 +7,7 @@
 + [8 String to Integer (atoi)(溢出检测）](algorithms/StringtoInteger)
 + [21 Merge Two Sorted Lists](algorithms/MergeTwoSortedLists)
 + [31 Next Permutation](algorithms/NextPermutation)
++ [60 Permutation Sequence](algorithms/PermutationSequence)
 + [65 Valid Number(细节题,状态机)](algorithms/ValidNumber)
 + [191 Number of 1 Bits(位运算)](algorithms/Numberof1Bits)
 + [198 House Robber(DP)](algorithms/HouseRobber)

algorithms/PermutationSequence/README.md

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+## 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.**
+
+## Solution
+
+求康托展开式，具体算法见[康托展开算法](https://github.com/krystism/algorithms/tree/master/cantor)
+
+由于是求第k个排列，而康托展开求的是根据比它小的数量，因此需要k减一
+
+## Code
+
+```c
+char *listRemove(char *s, int i)
+{
+	int n = strlen(s), j;
+	for (j = i; j < n - 1; ++j)
+		s[j] = s[j + 1];
+	s[j] = 0;
+	return s;
+}
+char *getPermutation(int n, int k) {
+	char *ans = malloc(sizeof(char) * (n + 1));
+	char *nums = malloc(sizeof(char) * n);
+	const int FAC[] = {1,1,2,6,24,120,720,5040,40320,362880};
+	for (int i = 0; i < n; ++i)
+		nums[i] = '0' + i + 1;
+	k--;
+	int cur = 0;
+	for (int i = n - 1; i > 0; --i) {
+		int index = k / FAC[i];
+		k %= FAC[i];
+		ans[cur++] = nums[index];
+		listRemove(nums, index);
+	}
+	ans[cur++] = nums[0];
+	ans[cur] = 0;
+	free(nums);
+	return ans;
+}
+```

algorithms/PermutationSequence/permutationSequence.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+char *listRemove(char *s, int i)
+{
+	int n = strlen(s), j;
+	for (j = i; j < n - 1; ++j)
+		s[j] = s[j + 1];
+	s[j] = 0;
+	return s;
+}
+char *getPermutation(int n, int k) {
+	char *ans = malloc(sizeof(char) * (n + 1));
+	char *nums = malloc(sizeof(char) * n);
+	const int FAC[] = {1,1,2,6,24,120,720,5040,40320,362880};
+	for (int i = 0; i < n; ++i)
+		nums[i] = '0' + i + 1;
+	k--;
+	int cur = 0;
+	for (int i = n - 1; i > 0; --i) {
+		int index = k / FAC[i];
+		k %= FAC[i];
+		ans[cur++] = nums[index];
+		listRemove(nums, index);
+	}
+	ans[cur++] = nums[0];
+	ans[cur] = 0;
+	free(nums);
+	return ans;
+}
+int main(int argc, char **argv)
+{
+	for (int i = 1; i <= 6; ++i)
+		printf("%s\n", getPermutation(3, i));
+	return 0;
+}