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:
- You must do this in-place without making a copy of the array.
- 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
- Python
- Java
- C++
# Move Zeroes
# 双指针
# Time Complexity: O(n), Space Complexity: O(1)
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
slow = 0
for fast in range(len(nums)):
if nums[fast] != 0:
nums[slow] = nums[fast]
slow += 1
for i in range(slow, len(nums)):
nums[i] = 0
// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
public void moveZeroes(int[] nums) {
int slow = 0;
for (int fast = 0; fast < nums.length; ++fast) {
if (nums[fast] != 0) {
nums[slow++] = nums[fast];
}
}
for (int i = slow; i < nums.length; ++i) {
nums[i] = 0;
}
}
}
// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int slow = 0;
for (int fast = 0; fast < nums.size(); ++fast) {
if (nums[fast] != 0) {
swap(nums[slow++], nums[fast]);
}
}
}
};
上面的代码实际上重复 实现了removeElement()，所以本题实际上可以直接调用removeElement()，只是最后要把后半截设置为 0。s
- Python
- Java
- C++
# Move Zeroes
# 双指针
# Time Complexity: O(n), Space Complexity: O(1)
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
slow = self.removeElement(nums, 0)
for i in range(slow, len(nums)):
nums[i] = 0
// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
public void moveZeroes(int[] nums) {
int slow = removeElement(nums, 0);
for (int i = slow; i < nums.length; ++i) {
nums[i] = 0;
}
}
}
// Move Zeroes
// 双指针
// Time Complexity: O(n), Space Complexity: O(1)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int slow = removeElement(nums, 0);
for (int i = slow; i < nums.size(); ++i) {
nums[i] = 0;
}
}
};