Kth Smallest Element in a BST
題意：
Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
Hint:
Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).
思路：
在一個 BST 中找第 k 小的數，而且利用 BST 的性質來做。找第 k 小什麼時候最方便呢？如果數組是有序的那遍歷到第 k 個就好了。
BST 可不可以變成一個有序的數組呢？或者說，什麼時候我們會從 BST 中得到一組有序的數呢？由 BST 左子樹的值都小於根節點，右子樹的值都大於根節點。所以中序遍歷的時候得到的就是有序數組。那麼，只要中序遍歷到第 k 個輸出就好了。
代碼：
/**
* 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:
int kthSmallest(TreeNode* root, int k) {
int cur = 0;
/* 非遞歸中序遍歷 */
stack<TreeNode*> s;
TreeNode* p = root;
while(p != NULL || !s.empty()) {
while(p != NULL) {
s.push(p);
p = p->left;
}
if(!s.empty()) {
p = s.top();
s.pop();
cur++;
/* 第 K 個 */
if(cur == k) return p->val;
p = p->right;
}
}
}
};Delete Node in a Linked List
題意：
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.
思路：
刪除給定的節點。用下個節點的值代替當前節點，並釋放下個節點空間。
代碼：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
ListNode* next = node->next;
*(node) = *(node->next);
delete next;
}
};Lowest Common Ancestor of a Binary Search Tree
題意：
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6.
Another example is LCA of nodes 2 and 4 is 2,
since a node can be a descendant of itself according to the LCA definition.
思路：
在二叉搜索樹上尋找兩個節點的最近公共祖先。利用二叉搜索樹的性質，用 p 和 q 的值與根節點的比較結果決定最近公共祖先的位置。
代碼：
/**
* 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* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* t;
/* 讓 p 比 q 小 */
if (p->val > q->val) {
t = p;
p = q;
q = t;
}
/* p 和 q 在根節點兩側，或者 p 和 q 某一個等於根節點，根節點就是最近公共祖先 */
if (p->val <= root->val && q->val >= root->val)
{
return root;
}
if(p->val < root->val && q->val < root->val)
{
/* p 和 q 都在根節點左側，最近公共祖先肯定在左側 */
return lowestCommonAncestor(root->left, p, q);
} else {
/* p 和 q 都在根節點右側，最近公共祖先肯定在右側 */
return lowestCommonAncestor(root->right, p, q);
}
}
};Valid Anagram
題意：
Given two strings s and t, write a function to determine if t is an anagram of s.
For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.
Note:
You may assume the string contains only lowercase alphabets.
思路：
一個單詞是否可以由另一個單詞重組而成。首先兩個單詞的長度要一樣，其次包含每個字母的個數要一樣。我的想法就是直接使用 map 統計字母的個數就好了。
代碼：
class Solution {
public:
bool isAnagram(string s, string t) {
if(s.size() != t.size()) return false;
map<char,int> letter;
for(int i = 0;i < s.size(); i++) {
letter[s[i]] ++;
}
for(int j = 0;j < t.size(); j++) {
if( --letter[t[j]] < 0) {
return false;
}
}
return true;
}
};Add Digits
題意：
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
思路：
沒有最後的要求的話，肯定是循環+遞歸，done。可是 O(1) 啊...有種直覺肯定是數學規律了...然而沒有想出來...
真正的 Wiki 在這：https://en.wikipedia.org/wiki/Digital_root 叫做 digital root. 規律爲：
也可以概括爲：
其實用了同餘的性質：
代碼：
遞歸的：
class Solution {
public:
int addDigits(int num) {
int ans = 0;
do {
ans += num % 10;
num /= 10;
} while (num);
if(ans / 10) return addDigits(ans);
else return ans;
}
};class Solution {
public:
int addDigits(int num) {
return 1 + ((num-1)%9);
}
};
