问题: 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.git
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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方法: 先以O(n)复杂度遍历将元素修改成平方,而后用快速排序的复杂度将数组进行排序,后续要优化为总体O(n)的复杂度。github
具体实现:数组
class SquaresOfASortedArray {
fun sortedSquares(A: IntArray): IntArray {
for (index in A.indices) {
A[index] *= A[index]
}
Arrays.sort(A)
return A
}
}
fun main(args: Array<String>) {
val input = intArrayOf(-4, -1, 0, 3, 10)
val squaresOfASortedArray = SquaresOfASortedArray()
CommonUtils.printArray(squaresOfASortedArray.sortedSquares(input).toTypedArray())
}
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有问题随时沟通bash