Given an array nums with n objects colored red, white, or blue, sort them in-placeso that objects of the same color are adjacent, with the colors in the order red, white, and blue.
We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.
You must solve this problem without using the library's sort function.
Example 1:
Input: nums = [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
Example 2:
Input: nums = [2,0,1]
Output: [0,1,2]
Constraints:
n == nums.length
1 <= n <= 300
nums[i] is either 0, 1, or 2.
Follow up: Could you come up with a one-pass algorithm using only constant extra space?
Solutions
Solution 1: Three Pointers
We define three pointers $i$, $j$, and $k$. Pointer $i$ is used to point to the rightmost boundary of the elements with a value of $0$ in the array, and pointer $j$ is used to point to the leftmost boundary of the elements with a value of $2$ in the array. Initially, $i=-1$, $j=n$. Pointer $k$ is used to point to the current element being traversed, initially $k=0$.
When $k < j$, we perform the following operations:
If $nums[k] = 0$, then swap it with $nums[i+1]$, then increment both $i$ and $k$ by $1$;
If $nums[k] = 2$, then swap it with $nums[j-1]$, then decrement $j$ by $1$;
If $nums[k] = 1$, then increment $k$ by $1$.
After the traversal, the elements in the array are divided into three parts: $[0,i]$, $[i+1,j-1]$ and $[j,n-1]$.
The time complexity is $O(n)$, where $n$ is the length of the array. Only one traversal of the array is needed. The space complexity is $O(1)$.
/** Do not return anything, modify nums in-place instead. */functionsortColors(nums:number[]):void{leti=-1;letj=nums.length;letk=0;while(k<j){if(nums[k]===0){++i;[nums[i],nums[k]]=[nums[k],nums[i]];++k;}elseif(nums[k]===2){--j;[nums[j],nums[k]]=[nums[k],nums[j]];}else{++k;}}}