[Leetcode] Sort Colors Color Sorting

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?

The nature of the problem is still a sort of problem, the topic gives the hint that you can use the count sort, need to traverse the array two times, then first look at this method, because there are only three different elements in the array, so it is easy to implement.

-First iterate over the original array, recording the number of 0,1,2 separately

-then update the original array, by the number of 0,1,2,

Solution One:

//Count SortclassSolution { Public:    voidSortcolors (intA[],intN) {intcount[3] = {0}, idx =0;  for(inti =0; I < n; ++i) + +Count[a[i]];  for(inti =0; I <3; ++i) { for(intj =0; J < Count[i]; ++j) {A[idx++] =i; }        }    }};

The problem is also to allow only one iteration of the array to solve, then I need to use a double pointer, respectively, from the original array to the center of the first movement.

-Define red pointer to start position, blue pointer to end position

-Traverse the original array from the beginning, and if you encounter 0, swap the value with the red pointer and move the red pointer back one bit. If 2 is encountered, the value and the value pointed to by the blue pointer are swapped and the blue pointer is moved forward one bit. If 1 is encountered, the traversal continues.

Solution Two:

classSolution { Public:    voidSortcolors (intA[],intN) {intRed =0, blue = n-1;  for(inti =0; I <= blue; ++i) {if(A[i] = =0) {swap (a[i], a[red++]); } Else if(A[i] = =2) {Swap (A[i--], a[blue--]); }         }    }};

[Leetcode] Sort Colors Color Sorting