Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
Example
Given 4 points: (1,2), (3,6), (0,0), (1,3).
The maximum number is 3.
LeeCode上的原题，可参见我之前的博客Invert Binary Tree 翻转二叉树。
解法一：
// Recursion class Solution { public: /** * @param root: a TreeNode, the root of the binary tree * @return: nothing */ void invertBinaryTree(TreeNode *root) { if (!root) return; TreeNode *tmp = root->left; root->left = root->right; root->right = tmp; invertBinaryTree(root->left); invertBinaryTree(root->right); } };
解法二：
// Non-Recursion class Solution { public: /** * @param root: a TreeNode, the root of the binary tree * @return: nothing */ void invertBinaryTree(TreeNode *root) { if (!root) return; queue<TreeNode*> q; q.push(root); while (!q.empty()) { TreeNode* node = q.front(); q.pop(); TreeNode *tmp = node->left; node->left = node->right; node->right = tmp; if (node->left) q.push(node->left); if (node->right) q.push(node->right); } } };
本文转自博客园Grandyang的博客，原文链接：翻转二叉树[LintCode] Invert Binary Tree ，如需转载请自行联系原博主。
