SquaresOfASortedArray.cpp

// Source : https://leetcode.com/problems/squares-of-a-sorted-array/
// Author : Hao Chen
// Date   : 2019-03-26

/***************************************************************************************************** 
 *
 * 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.
 * 
 ******************************************************************************************************/

class Solution {
public:
    vector<int> sortedSquares(vector<int>& A) {
        // find the place, negative numbers are right, positive number are right.
        // two pointer, one goes left, another goes right.

        //using binary search algorithm
        const int len = A.size();
        int low = 0, high = len- 1;
        int mid =0;
        while (low <= high) {
            mid = low + (high - low)/2;
            if (A[mid] >= 0 ) high = mid - 1;
            if (A[mid] < 0 ) low = mid + 1;
        }

        //TRICKY: make sure  A[mid] <= 0 or A[mid] is A[0]
        if (A[mid] > 0 && mid > 0 ) mid--;
        //cout << mid << " - "<< A[mid]<<  endl;

        vector<int> result;
        low = mid; high = mid+1;
        while ( low >=0 && high < len ) {
            if ( abs(A[low]) < abs(A[high]) ) {
                result.push_back(A[low] * A[low]);
                low --;
            }else {
                result.push_back(A[high] * A[high]);
                high++;
            }
        }

        for (;low >= 0; low--) result.push_back(A[low] * A[low]);
        for (;high<len; high++) result.push_back(A[high] * A[high] );

        return result;
    }
};