title: 905. Sort Array By Parity
tags:
- Two Pointers
- sorting
categories: leetcode
comments: false
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& nums) {
vector<int> ret;
int n = nums.size();
int j = n-1;
while(j>-1){
if(nums[j]%2==0) ret.push_back(nums[j]);
j--;
}
j = 0;
while(j < n){
if(nums[j]&1) ret.push_back(nums[j]);
j++;
}
return ret;
}
};
option 2 - one pass , two pointers
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& nums) {
int n = nums.size();
vector<int> ret(n,0);
int l=0, r = n-1;
for(int i=0;i<n;++i){
if(nums[i]%2==0) ret[l++] = nums[i];
else ret[r--] = nums[i];
}
return ret;
}
};
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& nums) {
int n = nums.size();
int l = 0, r = n-1;
while(l<r){
while(l<r && nums[l]%2==0) l++;
while(l<r && nums[r]%2!=0) r--;
swap(nums[l++], nums[r--]);
}
return nums;
}
};
- time complexity
O(n)
- space complexity
O(1)