Leetcode skimming ---977

subject ： Given a, press   Non decreasing order   Sorted array of integers  nums, return   The square of each number   A new array of , According to the requirements   Non decreasing order   Sort .

Input ：nums = [-4,-1,0,3,10]

Output ：[0,1,9,16,100]

Method 1 ： Direct sort

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

Complexity analysis

Time complexity ：O(nlogn)

Spatial complexity ：O(logn)

Method 2 ： Double pointer

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int n = nums.size();
        int negative = -1;
        for (int i = 0; i < n; ++i) {
            if (nums[i] < 0) {
                negative = i;
            } else {
                break;
            }
        }
        vector<int> ans;
        int i = negative, j = negative + 1;
        while(i >= 0 || j < n) {
            if (i < 0) {
                ans.push_back(nums[j] * nums[j]);
                ++j;
            } else if (j == n) {
                ans.push_back(nums[i] * nums[i]);
                --i;
            } else if (nums[i] * nums[i] < nums[j] * nums[j]) {
                ans.push_back(nums[i] * nums[i]);
                --i;
            } else {
                ans.push_back(nums[j] * nums[j]);
                ++j;
            }
        }
        return ans;
    }
};

Complexity analysis

Time complexity ：O(n)

Spatial complexity ：O(1)

Method 3 ： Double pointer

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0, j = n - 1, pos = n - 1; i <= j; ) {
            if (nums[i] * nums[i] > nums[j] * nums[j]) {
                ans[pos] = nums[i] * nums[i];
                ++i;
            } else {
                ans[pos] = nums[j] * nums[j];
                --j;
            }
            --pos;
        }
        return ans;
    }
};

Complexity analysis

Time complexity ：O(n)

Spatial complexity ：O(1)