LeetCode 75. Sort Colors（排序顏色）

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?

 

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題目標籤：Array

 

　　這道題目給了咱們一個顏色的array，讓咱們sort一下，按照0，1，2的順序。根據題目要求，咱們只能遍歷array一次，能夠用到two pointers來實現。設一個指針red 在開頭，blue 在最後。想法就是，遇到紅色0，就交換，把0放到最左邊去；遇到藍色2就交換，把2都放到最右邊去，這樣1就會被保留在最中間。須要注意的是，當把藍色2交換完畢以後，須要i--， 停留 i 在原地一次，由於還須要繼續檢查 被2交換回來的數字。那當遇到紅色0，交換完畢不須要停留i 的緣由是， 交換回來的只多是1，對於1，咱們不須要作任何處理，直接過就能夠。

 

 

 

Java Solution:

Runtime beats 55.91% 

完成日期：07/24/2017

關鍵詞：Array

關鍵點：用two pointers，一頭一尾放置紅色和藍色，保留白色在中間

 

 

 1 public class Solution 
 2 {
 3     public void sortColors(int[] nums) 
 4     {
 5         int red = 0;
 6         int blue = nums.length-1;
 7         
 8         for(int i=0; i<=blue; i++)
 9         {
10             if(nums[i] == 0) // if find 0, swap with red pointer
11             {
12                 int temp = nums[i];
13                 nums[i] = nums[red];
14                 nums[red] = temp;
15                 
16                 red++;
17             }
18             else if(nums[i] == 2) // if find 2, swap with blue pointer
19             {
20                 int temp = nums[i];
21                 nums[i] = nums[blue];
22                 nums[blue] = temp;
23                 
24                 i--;
25                 blue--;
26             }
27        
28         }
29     }
30 }

參考資料：

http://www.cnblogs.com/grandyang/p/4341243.html

 

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