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New Problem C++ Solution - "Squares of a Sorted Array"
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algorithms/cpp/squaresOfASortedArray/SquaresOfASortedArray.cpp
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// Source : https://leetcode.com/problems/squares-of-a-sorted-array/ | ||
// Author : Hao Chen | ||
// Date : 2019-03-26 | ||
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/***************************************************************************************************** | ||
* | ||
* 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. | ||
* | ||
******************************************************************************************************/ | ||
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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. | ||
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//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; | ||
} | ||
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//TRICKY: make sure A[mid] <= 0 or A[mid] is A[0] | ||
if (A[mid] > 0 && mid > 0 ) mid--; | ||
//cout << mid << " - "<< A[mid]<< endl; | ||
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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++; | ||
} | ||
} | ||
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for (;low >= 0; low--) result.push_back(A[low] * A[low]); | ||
for (;high<len; high++) result.push_back(A[high] * A[high] ); | ||
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return result; | ||
} | ||
}; |