linear-list/array/move-zeroes

Move Zeroes

描述

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

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

分析

这题跟 "Remove Element" 思路一模一样，只是最后要把后半截设置为 0。

代码

# Move Zeroes
# 双指针
# Time Complexity: O(n), Space Complexity: O(1)
class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        i = 0
        for j in range(len(nums)):
            if nums[j] != 0:
                nums[i] = nums[j]
                i += 1
        for j in range(i, len(nums)):
            nums[j] = 0

// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public void moveZeroes(int[] nums) {
        int i = 0;
        for (int j = 0; j < nums.length; ++j) {
            if (nums[j] != 0) {
                nums[i++] = nums[j];
            }
        }
        for (int j = i; j < nums.length; ++j) {
            nums[j] = 0;
        }
    }
}

// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int i = 0;
        for (int j = 0; j < nums.size(); ++j) {
            if (nums[j] != 0) {
                swap(nums[i++], nums[j]);
            }
        }
    }
};

相关题目

• Remove Element