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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 \(O(1)\).

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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

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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;
    }
}

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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;
            }
        }
    }
};

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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++
        }
    }
}

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/**
 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;
        }
    }
}

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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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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;
    }
}

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