905. Sort Array By Parity

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problem

solution

option 1 - two pass

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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

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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;
    }
};

option 3 - in-place

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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;
    }
};

analysis

  • time complexity O(n)
  • space complexity O(1)