Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.
Return any array that satisfies this condition.
Example 1:
Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation:
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Example 2:
Input: nums = [0]
Output: [0]
Constraints:
1 <= nums.length <= 5000
0 <= nums[i] <= 5000
Thought：
- Simple question and easy to do!
- First traverse once to find the output of even items
- After traversing once, find the count item output
AC：
/*
* @lc app=leetcode.cn id=905 lang=cpp
*
* [905] 按奇偶排序数组
*/
// @lc code=start
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& nums) {
vector<int> res;
for(auto num : nums)
{
if(num % 2 == 0)
{
res.push_back(num);
}
}
for(auto num : nums)
{
if(num % 2 == 1)
{
res.push_back(num);
}
}
return res;
}
};
// @lc code=end
After reading the official problem solution, it was consistent with idea 1, so
I looked at the other two solutions:
