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.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
这道题让我们将一个给定数组中所有的0都移到后面,把非零数前移,要求不能改变非零数的相对应的位置关系,而且不能拷贝额外的数组,那么只能用替换法in-place来做,需要用两个指针,一个不停的向后扫,找到非零位置,然后和前面那个指针交换位置即可,参见下面的代码:
解法一:
class Solution { public: void moveZeroes(vector<int>& nums) { for (int i = 0, j = 0; i < nums.size(); ++i) { if (nums[i]) { swap(nums[i], nums[j++]); } } } };
下面这种解法的思路跟上面的没啥区别,写法稍稍复杂了一点。。
解法二:
class Solution { public: void moveZeroes(vector<int>& nums) { int left = 0, right = 0; while (right < nums.size()) { if (nums[right]) { swap(nums[left++], nums[right]); } ++right; } } };
本文转自博客园Grandyang的博客,原文链接:移动零[LeetCode] Move Zeroes ,如需转载请自行联系原博主。