Saturday, February 20, 2016

M面经Prepare: Positive-Negative partitioning preserving order - neverlandly - 博客园


M面经Prepare: Positive-Negative partitioning preserving order - neverlandly - 博客园
Given an array which has n integers,it has both positive and negative integers.
Now you need sort this array in a special way.After that,the negative integers should 
in the front,and the positive integers should in the back.Also the relative position 
should not be changed.  eg. -1 1 3 -2 2 ans: -1 -2 1 3 2.

Time: O(NlogN), space: O(1)/O(logN) depends on iteration/recursion

This can be done in O(nlogn) using divide and conquer scheme. Before starting the algorithm, please see the following observation:
The basic idea of the algorithm is as follows:
1. We recursively 'sort' two smaller arrays of size n/2 (here 'sort' is defined in the question)
2. Then we spend \theta(n) time merging the two sorted smaller arrays with O(1) space complexity.
How to merge?
Suppose the two sorted smaller array is A and B. A1 denotes the negative part of A, and A2 denotes positive part of A. Similarly, B1 denotes the negative part of B, and B2 denotes positive part of B.
2.1. Compute the inverse of A2 (i.e., A2') in \theta(|A2|) time; compute the inverse of B1 (i.e., B1') in \theta(|B1|) time. [See observation; the total time is \theta(n) and space is O(1)]
Thus the array AB (i.e., A1A2B1B2) becomes A1A2'B1'B2.
2.2. Compute the inverse of A2'B1' (i.e., B1A2) in \theta(|A2|) time. [See observation; the total time is \theta(n) and space is O(1)]
Thus the array A1A2'B1'B2 becomes A1B1A2B2. We are done.
Time complexity analysis:
T(n) = 2T(n/2) + \theta(n) = O(nlogn)

 5     public void rearrange(int[] arr) {
 6         if (arr==null || arr.length==0) return;
 7         rearrange(arr, 0, arr.length-1);
 8     }
 9     
10     public void rearrange(int[] arr, int l, int r) {
11         if (l == r) return;
12         int m = (l+r)/2;
13         rearrange(arr, l, m);
14         rearrange(arr, m+1, r);
15         merge(arr, l, m+1, r);
16     }
17     
18     public void merge(int[] arr, int s1, int s2, int e) {
19         int findPos1 = s1, findPos2 = s2;
20         while (findPos1<s2 && arr[findPos1] < 0) findPos1++;
21         while (findPos2<=e && arr[findPos2] < 0) findPos2++;
22         reverse(arr, findPos1, s2-1);
23         reverse(arr, s2, findPos2-1);
24         reverse(arr, findPos1, findPos2-1);
25     }
26     
27     public void reverse(int[] arr, int start, int end) {
28         while (start < end) {
29             int temp = arr[start];
30             arr[start] = arr[end];
31             arr[end] = temp;
32             start++;
33             end--;
34         }
35     }
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