Leetcode-sort Colors

Leetcode-sort Colors

Given an array with n objects colored red, white or blue, sort them so, objects of the same color is Adjacen T, with the colors in the order red, white and blue.

Here, we'll use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note:
You is not a suppose to use the library's sort function for this problem.

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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?

classSolution { Public:    voidSortcolors (vector<int>&nums) {        intPos0 =-1, POS1 =-1, Pos2 =-1; inti =0;  while(i<nums.size ()) {            if(Nums[i] = =0) {nums[++POS2] =2; nums[++POS1] =1; nums[++POS0] =0; }            Else if(Nums[i] = =1) {nums[++POS2] =2; nums[++POS1] =1; }            Else if(Nums[i] = =2) {nums[++POS2] =2; } I++; }    }};

Analysis, it is obvious that you can use the bucket ordering method, but the problem requires the use of an algorithm that traverses only once.

Reference: http://www.cnblogs.com/felixfang/p/3680047.html

The specific idea is as follows: Record the end position of the 0,1,2 three numbers. Then the entire array to traverse, encountered 0 o'clock, the number of 0 plus one, that is, 0 at the end of the position plus one, and then the value of 0, then, the original 1 is changed to 0, the end of the natural 1 is also added one, and the assignment is 1, the same is true for 2.

If you encounter 1, you do not have to consider the 0 position, only 1 and 2 can be adjusted.

If 2 is encountered, only 2 of the position is considered.

Diagrams:

000011112222 0...

000001111222 2...

Leetcode-sort Colors