977. Squares of a Sorted Array
Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
Example 1:
Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
Example 2:
Input: nums = [-7,-3,2,3,11]
Output: [4,9,9,49,121]
Constraints:
- 1 <= nums.length <= 104
- -104 <= nums[i] <= 104
- nums is sorted in non-decreasing order.
Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?
Solution
class Solution {
public int[] sortedSquares(int[] nums) {
int n = nums.length;
int[] res = new int[n];
int j = n - 1;
int lo = 0;
int hi = n - 1;
while (lo <= hi) {
int left = nums[lo] * nums[lo];
int right = nums[hi] * nums[hi];
if (right > left) {
res[j] = right;
hi--;
j--;
} else {
lo++;
res[j] = left;
j--;
}
}
return res;
}
}
Solution 2021-11-18
class Solution {
public int[] sortedSquares(int[] nums) {
int n = nums.length;
int[] res = new int[n];
int lo = 0;
int hi = n - 1;
for (int index = n - 1; index >= 0; index--) {
if (nums[lo] * nums[lo] > nums[hi] * nums[hi]) {
res[index] = nums[lo] * nums[lo];
lo++;
} else {
res[index] = nums[hi] * nums[hi];
hi--;
}
}
return res;
}
}
Solution 2021-11-30
class Solution {
public int[] sortedSquares(int[] nums) {
int[] res = new int[nums.length];
int lo = 0;
int hi = nums.length - 1;
for (int k = nums.length - 1; k >= 0; k--) {
if (nums[lo] * nums[lo] < nums[hi] * nums[hi]) {
res[k] = nums[hi] * nums[hi];
hi--;
} else {
res[k] = nums[lo] * nums[lo];
lo++;
}
}
return res;
}
}