Js源码
Trie
208. Implement Trie (Prefix Tree)
Implement a trie with insert, search, and startsWith methods.
Example:
Trie trie = new Trie();
trie.insert("apple"); trie.search("apple"); // returns true trie.search("app"); // returns false trie.startsWith("app"); // returns true trie.insert("app"); trie.search("app"); // returns true
Note:
You may assume that all inputs are consist of lowercase letters a-z.
All inputs are guaranteed to be non-empty strings.
211. Add and Search Word - Data structure design
Design a data structure that supports the following two operations:
void addWord(word) bool search(word)
search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.
Example:
addWord("bad") addWord("dad") addWord("mad") search("pad") -> false search("bad") -> true search(".ad") -> true search("b..") -> true
Note: You may assume that all words are consist of lowercase letters a-z.
思考路程
1 利用208创建的trie树，修改下搜索条件即可
并查集
547. Friend Circles
There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
Example 1:
Input: [[1,1,0], [1,1,0], [0,0,1]] Output: 2 Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. The 2nd student himself is in a friend circle. So return 2.
Example 2:
Input: [[1,1,0], [1,1,1], [0,1,1]] Output: 1 Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
Note:
N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.
思考路程
1 类似海龟岛，可以用海龟岛的思路
2 并查集，采用并查,
心得： 1 在测试的时候，最后一个测试用例一直没通过，写代码的时候，多想想是不是自己预先设置了条件
线段树
307. Range Sum Query - Mutable
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
The update(i, val) function modifies nums by updating the element at index i to val.
Example:
Given nums = [1, 3, 5]
sumRange(0, 2) -> 9 update(1, 2) sumRange(0, 2) -> 8
Constraints:
The array is only modifiable by the update function.
You may assume the number of calls to update and sumRange function is distributed evenly.
0 <= i <= j <= nums.length - 1
思考路程
1
