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# 283.Move-Zeroes

## 题目描述

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.
Example:
Input: [0,1,0,3,12]
Output: [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.

## 代码

### Approach 1: Sapce Sub-Optimal

Complexity Analysis Space Complexity : O(n). Since we are creating the "ans" array to store results. Time Complexity: O(n). However, the total number of operations are sub-optimal. We can achieve the same result in less number of operations.
class Solution {
public void moveZeroes(int[] nums) {
if (nums == null || nums.length == 0) return;
int n = nums.length;
int numZeroes = 0;
for (int i = 0; i < n; i++) {
if (nums[i] == 0) {
numZeroes++;
}
}
List<Integer> ans = new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
if (nums[i] != 0) {
}
}
while (numZeroes--) {
}
for (int i = 0; i < n; i++) {
nums[i] = ans[i];
}
}
}

### Appraoch 2: Space Optimal, Operation Sub-Optimal

Complexity Analysis
Space Complexity : O(1). Only constant space is used.
Time Complexity: O(n). However, the total number of operations are still sub-optimal. The total operations (array writes) that code does is n (Total number of elements).
class Solution {
public void moveZeroes(int[] nums) {
if (nums == null || nums.length == 0) return;
int lastNonZeroFoundAt = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
nums[lastNonZeroFoundAt++] = nums[i];
}
}
for (int i = lastNonZeroFoundAt; i < nums.length; i++) {
nums[i] = 0;
}
}
}

### Approach 3: Optimal

1. 1.
All elements before the slow pointer (lastNonZeroFoundAt) are non-zeroes.
2. 2.
All elements between the current and slow pointer are zeroes.
Complexity Analysis
Space Complexity : O(1). Only constant space is used.
Time Complexity: O(n). However, the total number of operations are optimal. The total operations (array writes) that code does is Number of non-0 elements.This gives us a much better best-case (when most of the elements are 0) complexity than last solution. However, the worst-case (when all elements are non-0) complexity for both the algorithms is same.
class Solution {
public void moveZeroes(int[] nums) {
for (int lastNonZeroFoundAt = 0, cur = 0; cur < nums.length; cur++) {
if (nums[cur] != 0) {
swap(nums[lastNonZeroFoundAt++], nums[cur]);
}
}
}
}