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75. Sort Colors

Description

Given an array nums 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.

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 $O(1)$.

  • class Solution {
        public void sortColors(int[] nums) {
            int i = -1, j = nums.length, k = 0;
            while (k < j) {
                if (nums[k] == 0) {
                    swap(nums, ++i, k++);
                } else if (nums[k] == 2) {
                    swap(nums, --j, k);
                } else {
                    ++k;
                }
            }
        }
    
        private void swap(int[] nums, int i, int j) {
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    }
    
  • class Solution {
    public:
        void sortColors(vector<int>& nums) {
            int i = -1, j = nums.size(), k = 0;
            while (k < j) {
                if (nums[k] == 0) {
                    swap(nums[++i], nums[k++]);
                } else if (nums[k] == 2) {
                    swap(nums[--j], nums[k]);
                } else {
                    ++k;
                }
            }
        }
    };
    
  • class Solution:
        def sortColors(self, nums: List[int]) -> None:
            i, j, k = -1, len(nums), 0
            while k < j:
                if nums[k] == 0:
                    i += 1
                    nums[i], nums[k] = nums[k], nums[i]
                    k += 1
                elif nums[k] == 2:
                    j -= 1
                    nums[j], nums[k] = nums[k], nums[j]
                else:
                    k += 1
    
    
  • func sortColors(nums []int) {
    	i, j, k := -1, len(nums), 0
    	for k < j {
    		if nums[k] == 0 {
    			i++
    			nums[i], nums[k] = nums[k], nums[i]
    			k++
    		} else if nums[k] == 2 {
    			j--
    			nums[j], nums[k] = nums[k], nums[j]
    		} else {
    			k++
    		}
    	}
    }
    
  • /**
     Do not return anything, modify nums in-place instead.
     */
    function sortColors(nums: number[]): void {
        let i = -1;
        let j = nums.length;
        let k = 0;
        while (k < j) {
            if (nums[k] === 0) {
                ++i;
                [nums[i], nums[k]] = [nums[k], nums[i]];
                ++k;
            } else if (nums[k] === 2) {
                --j;
                [nums[j], nums[k]] = [nums[k], nums[j]];
            } else {
                ++k;
            }
        }
    }
    
    
  • public class Solution {
        public void SortColors(int[] nums) {
            int i = -1, j = nums.Length, k = 0;
            while (k < j) {
                if (nums[k] == 0) {
                    swap(nums, ++i, k++);
                } else if (nums[k] == 2) {
                    swap(nums, --j, k);
                } else {
                    ++k;
                }
            }
        }
    
        private void swap(int[] nums, int i, int j) {
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    }
    
  • impl Solution {
        pub fn sort_colors(nums: &mut Vec<i32>) {
            let mut i = -1;
            let mut j = nums.len();
            let mut k = 0;
            while k < j {
                if nums[k] == 0 {
                    i += 1;
                    nums.swap(i as usize, k as usize);
                    k += 1;
                } else if nums[k] == 2 {
                    j -= 1;
                    nums.swap(j, k);
                } else {
                    k += 1;
                }
            }
        }
    }
    
    

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