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.
Approach #1 Bubble sort,我的解法,慢
class Solution {
public void moveZeroes(int[] nums) {
for(int i=0;i<nums.length;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i]==0&&j<nums.length){
int temp=nums[i];
nums[i]=nums[j];
nums[j]=temp;
}
}
}
}
}
时间复杂度O(N^2)
空间复杂度O(1)
改进:
public void moveZeroes(int[] nums) {
int j = 0;
for(int i = 0; i < nums.length; i++) {
if(nums[i] != 0) {
int temp = nums[j];
nums[j] = nums[i];
nums[i] = temp;
j++;
}
}
}
}
Approach #2 Insert Index
// Shift non-zero values as far forward as possible
// Fill remaining space with zeros
public void moveZeroes(int[] nums) {
if (nums == null || nums.length == 0) return;
int insertPos = 0;
for (int num: nums) {
if (num != 0) nums[insertPos++] = num;
}
while (insertPos < nums.length) {
nums[insertPos++] = 0;
}
}
先把非零元素放到应该在的地方,再根据array长度算出结尾需要补的零,补在后面即可。
时间复杂度O(N)
空间复杂度O(1)
Approach #3 Snowball
class Solution {
public void moveZeroes(int[] nums) {
int snowBallSize = 0;
for (int i=0;i<nums.length;i++){
if (nums[i]==0){
snowBallSize++;
}
else if (snowBallSize > 0) {
int t = nums[i];
nums[i]=0;
nums[i-snowBallSize]=t;
}
}
}
}
像滚雪球一样,把0 都滚到一起,然后最左侧的零和非零元素对调。
时间复杂度O(N)
空间复杂度O(1)