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?
使用两个变量来，标记0和1的位置。遍历一遍数组，即可。public void sortColors(int[] A) { if (A != null && A.length != 0 && A.length != 1) { int len = A.length; int left = 0; int right = len - 1; for (int i = 0; i < len;) { if (A[i] == 0 && i > left) { int temp = A[left]; A[left] = A[i]; A[i] = temp; left++; } else if (A[i] == 2 && right > i) { int temp = A[right]; A[right] = A[i]; A[i] = temp; right--; } else { i++; } } } }
很巧妙。public void sortColors(int[] A) { if (A != null && A.length != 0 && A.length != 1) { int i = 0; int j = 0; int k = 0; for (int m = 0; m < A.length; m++) { if (A[m] == 0) { A[i++] = 2; A[j++] = 1; A[k++] = 0; } else if (A[m] == 1) { A[i++] = 2; A[j++] = 1; } else if (A[m] == 2) { A[i++] = 2; } } } }
原文地址：http://blog.csdn.net/u010378705/article/details/29581797
