LeetCode: 283 Move Zeroes(easy)

Posted 2020-10-05 玲珑子

题目：

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:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

代码：

不知道为什么，在本地可以运行，提交上去显示 Submission Result: Runtime Error ，感觉可能是越界？不知道。。调不出来了。。（把0和后面非0的交换）

 1 class Solution {
 2 public:
 3     vector<int> moveZeroes(vector<int>& nums) {
 4         for(auto zero = nums.begin(); zero < nums.end(); zero++){
 5             if ( *zero == 0 ){
 6                 auto notzero = zero;
 7                 for(notzero; notzero < nums.end(); notzero++){
 8                     if(*notzero != 0)
 9                         break;
10                 }  
11　　　　　　　　　if (zero != notzero)
12                    iter_swap(zero, notzero);
13             }
14         }
15         return nums;
16     }
17 };

换个思路（把后面非0的元素和前面的0元素交换）：

 1 class Solution {
 2 public:
 3     void moveZeroes(vector<int>& nums) {
 4         int last = 0, cur = 0;
 5         while(cur < nums.size()) {
 6             if(nums[cur] != 0) {
 7                 swap(nums[last], nums[cur]);
 8                 last++;
 9             }
10             cur++;
11         }  
12     }
13 };

别人的代码：

 1 class Solution {
 2 public:
 3     void moveZeroes(vector<int>& nums) {
 4         int nonzero = 0;
 5         for(int n:nums) {
 6             if(n != 0) {
 7                 nums[nonzero++] = n;
 8             }
 9         }
10         for(;nonzero<nums.size(); nonzero++) {
11             nums[nonzero] = 0;
12         }
13     }
14 };