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.
click to show follow up.
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?
算法一:紅白藍三指針從前向後移動
這個算法理解可以借用刷牆。
假設有x個紅,y個白,z個藍
先全部刷成藍(x+y+z)。
再從頭刷(x+y)個成白。
再從頭刷x個成紅。
class Solution { public: void sortColors(int A[], int n) { int red = -1, white = -1, blue = -1; for (int i=0; i
算法參考自:
https://leetcode.com/discuss/1827/anyone-with-one-pass-and-constant-space-solution
算法二:紅白二指針從前向後, 藍指針從後往前移動
class Solution { public: void sortColors(int A[], int n) { int red = -1, white = 0, blue = n; while (white < blue) { if (A[white] == 1) { ++white; } else if (A[white] == 0) { A[white++] = 1; A[++red] = 0; } else { A[white] = A[--blue]; A[blue] = 2; } } } };