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283. Move Zeroes

Description

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

 

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

  • 1 <= nums.length <= 104
  • -231 <= nums[i] <= 231 - 1

 

Follow up: Could you minimize the total number of operations done?

Solutions

Solution 1: Two Pointers

Thinking

Zeros must go to the end while nonzeros keep their order. Let \(k\) be the next slot for a nonzero.

Each nonzero is swapped with \(nums[k]\) and \(k\) advances, so the prefix of length \(k\) is the original nonzero sequence.

We use a pointer \(k\) to record the current position to insert, initially \(k = 0\).

Then we iterate through the array \(\textit{nums}\), and each time we encounter a non-zero number, we swap it with \(\textit{nums}[k]\) and increment \(k\) by 1.

This way, we can ensure that the first \(k\) elements of \(\textit{nums}\) are non-zero, and their relative order is the same as in the original array.

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).

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class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        k = 0
        for i, x in enumerate(nums):
            if x:
                nums[k], nums[i] = nums[i], nums[k]
                k += 1

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class Solution {
    public void moveZeroes(int[] nums) {
        int k = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] != 0) {
                int t = nums[i];
                nums[i] = nums[k];
                nums[k++] = t;
            }
        }
    }
}

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class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int k = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            if (nums[i]) {
                swap(nums[i], nums[k++]);
            }
        }
    }
};

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func moveZeroes(nums []int) {
    k := 0
    for i, x := range nums {
        if x != 0 {
            nums[i], nums[k] = nums[k], nums[i]
            k++
        }
    }
}

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/**
 Do not return anything, modify nums in-place instead.
 */
function moveZeroes(nums: number[]): void {
    let k = 0;
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i]) {
            [nums[i], nums[k]] = [nums[k], nums[i]];
            ++k;
        }
    }
}

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impl Solution {
    pub fn move_zeroes(nums: &mut Vec<i32>) {
        let mut k = 0;
        let n = nums.len();
        for i in 0..n {
            if nums[i] != 0 {
                nums.swap(i, k);
                k += 1;
            }
        }
    }
}

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/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var moveZeroes = function (nums) {
    let k = 0;
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i]) {
            [nums[i], nums[k]] = [nums[k], nums[i]];
            ++k;
        }
    }
};

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void moveZeroes(int* nums, int numsSize) {
    int k = 0;
    for (int i = 0; i < numsSize; ++i) {
        if (nums[i] != 0) {
            int t = nums[i];
            nums[i] = nums[k];
            nums[k++] = t;
        }
    }
}

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