977. Squares of a Sorted Array

發表於 | 分類於

problem

solution

option 1

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class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        for(int & a:nums) a=a*a;
        sort(nums.begin(), nums.end());
        return nums;
    }
};

option 2 - Two Pointers

利用雙索引，去比較左右索引對應到的數字的平方，比較大小。

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class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int n = nums.size(), l=0, r = n-1;
        vector<int>ret(n,0);
        for(int i=n-1;i>-1;--i){
            if(nums[l]*nums[l] > nums[r]*nums[r]){
                ret[i] = nums[l]*nums[l];
                l++;
            }
            else{
                ret[i] = nums[r]* nums[r];
                r--;
            }
        }
        return ret;
    }
};
  • other version
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    vector<int> ret(nums.size(),0);
    int l = 0,r = nums.size()-1;
    int k =nums.size()-1;
    while(l<=r){
        if(abs(nums[l])>=abs(nums[r])) ret[k--] =pow(nums[l++],2);
        else{
            ret[k--] = pow(nums[r--],2);
        }
    }
    return ret;

analysis

  • option 1
    • time complexity O(nlogn)
    • space complexity O(1)
  • option 2
    • time complexity O(n)
    • space complexity O(1)