Leetcode solution 75. Sort Colors

Problem Statement

Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library's sort function for this problem.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Follow up:

A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
Could you come up with a one-pass algorithm using only constant space?

Problem link

Video Tutorial

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Thought Process

This is the classic Dutch National Flag Problem.

Two pass with constant space is easy with count sort. (Reminds me of Radix Sort somehow) An extension would be what happens if there are K elements (K >=3)

If do it in one pass, then the idea must involve pointers and swap. Start from left to right, if 1, pass; if 0, move to the left; if 2, move to the right

Solutions

Two pass

 1 public void sortColorsCountingSort(int[] A) {
 2     if (A == null || A.length == 0) return;
 3 
 4     int[] buckets = new int[3];
 5 
 6     for (int i = 0; i < A.length; i++) {
 7         buckets[A[i]] += 1;
 8     }
 9 
10     int index = 0;
11     for (int i = 0; i < 3; i++) {
12         for (int j = 0; j < buckets[i]; j++) {
13             A[index] = i;
14             index++;
15         }
16     }
17 }

Time Complexity: O(N), where N is the array size, N + 3*N = 4N = O(N)

Space Complexity: O(1) since only need 3 elements in this case, still considered constant

One Pass

 1 public void sortColors(int[] A) {
 2     if (A == null || A.length == 0) return;
 3 
 4     int left = 0;
 5     int right = A.length - 1;
 6 
 7     int i = 0; 
 8     while (i <= right) { // has to be <=, since cur value still needs to be evaluated if swapped back
 9         if (A[i] == 1) {
10             i++;
11         } else if (A[i] == 0) {
12             swap(A, i, left);
13             left++;
14             i++; // don't forget this one, we can only exchange 0 or 1 back here, so move forward no matter what
15         } else { // A[i] == 2
16             swap(A, i, right);
17             right--;
18         }
19     }
20 }

Time Complexity: O(N), where N is the array size

Space Complexity: O(1)

References

Leetcode official solution

来源：https://www.cnblogs.com/baozitraining/p/12501620.html

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