Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number, also in sorted non-decreasing order.

 

Example 1:

Input: [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Example 2:

Input: [-7,-3,2,3,11]
Output: [4,9,9,49,121]
 

Note:

1 <= A.length <= 10000
-10000 <= A[i] <= 10000
A is sorted in non-decreasing order.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/squares-of-a-sorted-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

跟刚才那道题很像，找到一个神奇思路，两边的平方一定是最大的，so~双指针从两边往中间移动就很简单了，一开始从中间往两边走就麻烦很多

但是这个时间速度很慢~百分之23%

class Solution {
public:
    vector<int> sortedSquares(vector<int>& A) {
        int n=A.size();
        vector<int> num(n);
        if(n==0){
            return num;
        }
        int a=0;
        int b=n-1;
        int na,nb;
        while(n>0&&a!=b){
            n--;
            na=A[a]*A[a];
            nb=A[b]*A[b];
            if(na>nb){
                a++;
                num[n]=na;
            }else{
                b--;
                num[n]=nb;
            }
        }
        if(n>0) num[0]=A[a]*A[a];
        return num;
    }
};

暴力的老哥都比我快

class Solution {
public:
    vector<int> sortedSquares(vector<int>& A) {
        //暴力求解 双指针
        int start =0,end=A.size()-1;
        vector<int> res;
        while(start<=end){
            if(abs(A[start])>abs(A[end])){
                res.push_back(A[start]*A[start]);
                start++;
            }else{
                res.push_back(A[end]*A[end]);
                end--;
            }
        }
        reverse(res.begin(),res.end());
        return res;
    }
};