Leetcode: Move Zeroes

最新推荐文章于 2022-09-09 22:57:00 发布

转载最新推荐文章于 2022-09-09 22:57:00 发布 · 54 阅读

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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.

第二遍做法：双指针压缩法：O(N), 空间：O(1)

实际上就是将所有的非0数向前尽可能的压缩，最后把没压缩的那部分全置0就行了。比如103040，先压缩成134，剩余的3为全置为0。过程中需要一个指针记录压缩到的位置。

 1 public class Solution {
 2     public void moveZeroes(int[] nums) {
 3         if (nums==null || nums.length==0) return;
 4         int cur=0;
 5         for (int i=0; i<nums.length; i++) {
 6             if (nums[i]!=0) {
 7                 nums[cur++] = nums[i];
 8             }
 9         }
10         for (int i=cur; i<nums.length; i++) {
11             nums[i] = 0;
12         }
13     }
14 }

Q：如果要把所有的0放在前面而不是后面呢？
A：同样的解题思路，但是是从后向前遍历，将非0数字压缩到后面。

双指针同向法：O(N^2),不是太好

Two pointer l,r,

l is to find each 0 in the list, r is to start from l, find first non-0

l, r should not both start from 0, because if r is to the left of l, we don't want to swap it. example: [1, 0]

 1 public class Solution {
 2     public void moveZeroes(int[] nums) {
 3         if (nums==null && nums.length==0) return;
 4         int l=0, r=0;
 5         while (l < nums.length && r < nums.length) {
 6             while (l<nums.length && nums[l]!=0) {
 7                 l++;
 8             }
 9             r = l;
10             while (r<nums.length && nums[r]==0) {
11                 r++;
12             }
13             if (l == nums.length || r == nums.length) break;
14             swap(nums, l, r);
15         }
16     }
17     
18     public void swap(int[] nums, int l, int r) {
19         int temp = nums[l];
20         nums[l] = nums[r];
21         nums[r] = temp;
22     }
23 }