leetcode_question_75 Sort Colors
2013-09-25 11:57
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Given an array with n objects colored red, white or blue, sort them 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.
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 an one-pass algorithm using only constant space?
Counting sort:
one-pass:
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.
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 an one-pass algorithm using only constant space?
Counting sort:
void sortColors(int A[], int n) { // Start typing your C/C++ solution below // DO NOT write int main() function if(A == NULL || n < 2)return; int num0 = 0, num1 = 0, num2 = 0; for(int i = 0; i < n; ++i) { if(A[i]==0)num0++; else if(A[i]==1)num1++; else num2++; } int i = 0; while(num0--) A[i++] = 0; while(num1--) A[i++] = 1; while(num2--) A[i++] = 2; }
one-pass:
void sortColors(int A[], int n) { // Start typing your C/C++ solution below // DO NOT write int main() function if(A == NULL || n < 2)return; int red = 0, blue = n-1; for(int i = red; i <= blue; ++i) { if(A[i] == 0) swap(A[red++],A[i]); else if(A[i] == 2) swap(A[i--],A[blue--]); } }
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