本文主要是介绍leetcode -- 75. Sort Colors，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目描述
题目难度：Medium
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?
AC代码
方法1：分别记录0、1、2这三个数字出现的次数，然后按照顺序写进数组；
方法2：先把0、1看作一类，2看作另一类，总共两类，遍历数组，让2全部都处于数组的右端；然后剩下的数组里只有0、1两类，按照上述步骤即可完成。三类不好分，但是两类很好分。这种思维值得学习。
class Solution {public void sortColors(int[] nums) {if(nums == null || nums.length <= 1) return;swapZeroAndTwo(nums);int i = nums.length - 1;while(i >= 0 && nums[i] == 2) i--;if(i <= 0) return;swapZeroAndOne(nums, i);}private void swapZeroAndTwo(int[] nums){int low = 0, high = nums.length - 1;while(low < high){if(nums[low] != 2) low++;else if(nums[high] == 2) high--;else{int tmp = nums[low];nums[low] = nums[high];nums[high] = tmp;}}}private void swapZeroAndOne(int[] nums, int end){int low = 0, high = end;while(low < high){if(nums[low] != 1) low++;else if(nums[high] == 1) high--;else{int tmp = nums[low];nums[low] = nums[high];nums[high] = tmp;}}}
}
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