Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

Solution: 2 Passes

First pass: mirror around diagonal
Second pass: mirror around y axis

Time complexity: O(n^2)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  void rotate(vector<vector<int>>& matrix) {
    const int n = matrix.size();
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j)
        swap(matrix[i][j], matrix[j][i]);
    for (int i = 0; i < n; ++i)
      reverse(begin(matrix[i]), end(matrix[i]));
  }
};

The post 花花酱 LeetCode 48. Rotate Image appeared first on Huahua's Tech Road.

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

class Solution {
public:
  bool isSymmetric(TreeNode* root) {
    return isMirror(root, root);
  }
private:
  bool isMirror(TreeNode* root1, TreeNode* root2) {
    if (!root1 && !root2) return true;
    if (!root1 || !root2) return false;
    return root1->val == root2->val 
           && isMirror(root1->right, root2->left) 
           && isMirror(root1->left, root2->right);
  }
};

"""
Author: Huahua
Running time: 48 ms
"""
class Solution:
  def isSymmetric(self, root):
    def isMirror(root1, root2):
      if not root1 and not root2: return True
      if not root1 or not root2: return False
      return root1.val == root2.val \
        and isMirror(root1.left, root2.right) \
        and isMirror(root2.left, root1.right)
    return isMirror(root, root)

Related Problems

• 花花酱 Leetcode 100. Same Tree

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