283. Move Zeroes

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LeetCode

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Problem

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 <= 10^{4}
• -2^{31} <= nums[i] <= 2^{31} - 1

Code

当有很多 0 的时候，下面的解法更好

class Solution {
    public void moveZeroes(int[] nums) {
        int p = 0;
        for(int i = 0; i < nums.length; i++){
            if (nums[i] != 0){
                // 排除[1,2,3,0,0,0]这种情况
                if(p != i) {
                    nums[p] = nums[i];
                    nums[i] = 0;
                }

                p++;
            }
        }
    }
}

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

当有很多 0 的时候，下面的解法更好

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        index = 0
        for i in range(0, len(nums)):
            if nums[i] != 0:
                if i != index:
                    nums[index] = nums[i]
                    nums[i] = 0
                index += 1

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        index = 0
        for i in range(0, len(nums)):
            if nums[i] != 0:
                nums[index] = nums[i]
                index += 1

        for i in range(index, len(nums)):
            nums[i] = 0