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What hurts more, the pain of hard work or the pain of regret?

Invert a binary tree

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew),
but you can’t invert a binary tree on a whiteboard so fuck off.

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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root == NULL) return root;
TreeNode* temp;
temp = invertTree(root->left);
root->left = invertTree(root->right);
root->right = temp;
return root;
}
};

problem

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input:(2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

分析(不用英文装逼了)
这个代码是抄来的,链接原作是这位大大。

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Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

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