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 for n = 3:

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

Given n and k, return the kth permutation sequence.

思路： 通过给定的n和k，不断的递减n，获取每一次的数组的索引值及k的新值，直到n为0，终止循环 将获取的每一次的索引值，从数组（数据每一次取值后，都需要删除该值）内取出值即可。

int* copyIntList(int* list, int length) {
	int* result = (int*)malloc(length * sizeof(int));

	for (int i = 0; i < length; i++) {
		result[i] = list[i];
	}

	return result;
}

int getIndex(int n, int k) {
	return (double)ceil((double)k / getFabric(n - 1)) - 1;
}

int getFabric(int n) {
	if (n == 0) return 1;
	int total = 1;
	for (int i = 1; i <= n; i++) {
		total *= i;
	}
	return total;
}

void deleteElementByIndex(int* list, int length, int position) {
	for (int i = position; i < length; i++) {
		list[i - 1] = list[i];
	}
}

char* getPermutationSequenceByIndex(int* list, int n, int k) {
	int _n = n, _k = k;
	int* _list = copyIntList(list, n);

	char* result = (char*)malloc((n + 1) * sizeof(char));
	result[n] = '\0';
	int curResultIndex = 0;

	while (_n > 0) {
		int curIndex = (int)getIndex(_n, _k);
		//dosomething 1.get index from list , then remove the index val of list;
		result[curResultIndex] = '0' + _list[curIndex];
		curResultIndex++;
		deleteElementByIndex(_list, _n, curIndex + 1);

		int nextLen = getFabric(_n - 1);
		int nextK = _k % nextLen == 0 ? nextLen : _k % nextLen;
		_k = _k <= nextLen ? _k : nextK;
		_n--;
	}

	free(_list);
	return result;
}

char* getPermutation(int n, int k) {
	//生成int动态数组
	int* p = (int*)malloc(n * sizeof(int));
	for (int i = 0; i < n; i++) {
		p[i] = i + 1;
	}
	char* result = getPermutationSequenceByIndex(p, n, k);
	free(p);
	return result;
}