An in-place algorithm for String Transformation

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本文介绍了一种高效的字符串处理算法，该算法能在O(n)的时间复杂度内将字符串中的偶数位置字符移动到字符串末尾，同时保持字符的相对顺序不变。通过详细步骤分解和示例说明，读者可以了解如何实现这一算法。

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不好意思，我没有进行翻译，觉得原文讲的很好：

Given a string, move all even positioned elements to end of string. While moving elements, keep the relative order of all even positioned and odd positioned elements same. For example, if the given string is “a1b2c3d4e5f6g7h8i9j1k2l3m4″, convert it to “abcdefghijklm1234567891234″ in-place and in O(n) time complexity.
Below are the steps:
1. Cut out the largest prefix sub-string of size of the form 3^k + 1. In this step, we find the largest non-negative integer k such that 3^k+1 is smaller than or equal to n (length of string)
2. Apply cycle leader iteration algorithm ( it has been discussed below ), starting with index 1, 3, 9…… to this sub-string. Cycle leader iteration algorithm moves all the items of this sub-string to their correct positions, i.e. all the alphabets are shifted to the left half of the sub-string and all the digits are shifted to the right half of this sub-string.
3. Process the remaining sub-string recursively using steps#1 and #2.
4. Now, we only need to join the processed sub-strings together. Start from any end ( say from left ), pick two sub-strings and apply the below steps:
….4.1 Reverse the second half of first sub-string.
….4.2 Reverse the first half of second sub-string.
….4.3 Reverse the second half of first sub-string and first half of second sub-string together.
5. Repeat step#4 until all sub-strings are joined. It is similar to k-way merging where first sub-string is joined with second. The resultant is merged with third and so on.
Let us understand it with an example:
Please note that we have used values like 10, 11 12 in the below example. Consider these values as single characters only. These values are used for better readability.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
a 1 b 2 c 3 d 4 e 5 f 6 g 7 h 8 i 9 j 10 k 11 l 12 m 13
After breaking into size of the form 3^k + 1, two sub-strings are formed of size 10 each. The third sub-string is formed of size 4 and the fourth sub-string is formed of size 2.
0 1 2 3 4 5 6 7 8 9
a 1 b 2 c 3 d 4 e 5

10 11 12 13 14 15 16 17 18 19
f 6 g 7 h 8 i 9 j 10
20 21 22 23
k 11 l 12
24 25
m 13
After applying cycle leader iteration algorithm to first sub-string:
0 1 2 3 4 5 6 7 8 9
a b c d e 1 2 3 4 5
10 11 12 13 14 15 16 17 18 19
f 6 g 7 h 8 i 9 j 10
20 21 22 23
k 11 l 12
24 25
m 13
After applying cycle leader iteration algorithm to second sub-string:
0 1 2 3 4 5 6 7 8 9
a b c d e 1 2 3 4 5
10 11 12 13 14 15 16 17 18 19
f g h i j 6 7 8 9 10
20 21 22 23
k 11 l 12
24 25
m 13
After applying cycle leader iteration algorithm to third sub-string:
0 1 2 3 4 5 6 7 8 9
a b c d e 1 2 3 4 5
10 11 12 13 14 15 16 17 18 19
f g h i j 6 7 8 9 10
20 21 22 23
k l 11 12
24 25
m 13
After applying cycle leader iteration algorithm to fourth sub-string:
0 1 2 3 4 5 6 7 8 9
a b c d e 1 2 3 4 5
10 11 12 13 14 15 16 17 18 19
f g h i j 6 7 8 9 10
20 21 22 23
k l 11 12
24 25
m 13
Joining first sub-string and second sub-string:
1. Second half of first sub-string and first half of second sub-string reversed.
0 1 2 3 4 5 6 7 8 9
a b c d e 5 4 3 2 1 <--------- First Sub-string
10 11 12 13 14 15 16 17 18 19
j i h g f 6 7 8 9 10 <--------- Second Sub-string
20 21 22 23
k l 11 12
24 25
m 13
2. Second half of first sub-string and first half of second sub-string reversed together( They are merged, i.e. there are only three sub-strings now ).
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
a b c d e f g h i j 1 2 3 4 5 6 7 8 9 10
20 21 22 23
k l 11 12
24 25
m 13
Joining first sub-string and second sub-string:
1. Second half of first sub-string and first half of second sub-string reversed.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
a b c d e f g h i j 10 9 8 7 6 5 4 3 2 1 <--------- First Sub-string
20 21 22 23
l k 11 12 <--------- Second Sub-string
24 25
m 13
2. Second half of first sub-string and first half of second sub-string reversed together( They are merged, i.e. there are only two sub-strings now ).
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
a b c d e f g h i j k l 1 2 3 4 5 6 7 8 9 10 11 12
24 25
m 13
Joining first sub-string and second sub-string:
1. Second half of first sub-string and first half of second sub-string reversed.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
a b c d e f g h i j k l 12 11 10 9 8 7 6 5 4 3 2 1 <----- First Sub-string
24 25
m 13 <----- Second Sub-string
2. Second half of first sub-string and first half of second sub-string reversed together( They are merged, i.e. there is only one sub-string now ).
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13
Since all sub-strings have been joined together, we are done.
How does cycle leader iteration algorithm work?
Let us understand it with an example:
Input:
0 1 2 3 4 5 6 7 8 9
a 1 b 2 c 3 d 4 e 5
Output:
0 1 2 3 4 5 6 7 8 9
a b c d e 1 2 3 4 5
Old index New index
0 0
1 5
2 1
3 6
4 2
5 7
6 3
7 8
8 4
9 9
Let len be the length of the string. If we observe carefully, we find that the new index is given by below formula:
if( oldIndex is odd )
newIndex = len / 2 + oldIndex / 2;
else
newIndex = oldIndex / 2;
So, the problem reduces to shifting the elements to new indexes based on the above formula.
Cycle leader iteration algorithm will be applied starting from the indices of the form 3^k, starting with k = 0.
Below are the steps:
1. Find new position for item at position i. Before putting this item at new position, keep the back-up of element at new position. Now, put the item at new position.
2. Repeat step#1 for new position until a cycle is completed, i.e. until the procedure comes back to the starting position.
3. Apply cycle leader iteration algorithm to the next index of the form 3^k. Repeat this step until 3^k < len.
Consider input array of size 28:
The first cycle leader iteration, starting with index 1:
1->14->7->17->22->11->19->23->25->26->13->20->10->5->16->8->4->2->1
The second cycle leader iteration, starting with index 3:
3->15->21->24->12->6->3
The third cycle leader iteration, starting with index 9:
9->18->9
Based on the above algorithm, below is the code:

#include<iostream>
#include<string>
#include<cmath>
using namespace std;

void swap(char &a, char &b) {
	b ^= a;
	a ^= b;
	b ^= a;
}

void reverse(char *s, int start, int end) {
	while (start < end) {
		swap(*(s + start), *(s + end));
		start++;
		end--;
	}
}

void cycleLeader(char *s, int offset, int len) {
	int i, index;
	char item;
	for (i = 1; i < len; i *= 3) {
		index = i;
		item = s[index + offset];
		do {
			if (index & 0x01)
				index = len / 2 + index / 2;
			else
				index = index / 2;
			swap(*(s + index + offset), item);
		} while (index != i);
	}
}

void inPlaceMove(char *s, int length) {
	int left_len = length;
	int part_len = 0;
	int k = 0;
	int offset = 0;
	while (left_len) {
		k = 0;
		while (pow((double)3, k) + 1 <= left_len)
			k++;
		part_len = pow((double)3, k - 1) + 1;
		left_len -= part_len;
		cycleLeader(s, offset, part_len);
		if (offset) {
			reverse(s, offset / 2, offset - 1);
			reverse(s, offset, offset + part_len / 2 - 1);
			reverse(s, offset / 2, offset + part_len / 2 - 1);
		}
		offset += part_len;
	}
}

int main(int argc, char *argv[]) {
	char s[] = "1a2b3c4d5e6f7g8h9i";
	int n = strlen(s);
	inPlaceMove(s, n);
	cout << s << endl;
	cin.get();
	return 0;
}