Trie Service
Trie
Medium
Description
Build tries from a list of <word, freq> pairs. Save top 10 for each node.
Example
Example1
Input:
<"abc", 2>
<"ac", 4>
<"ab", 9>
Output:<a[9,4,2]<b[9,2]<c[2]<>>c[4]<>>>
Explanation:
Root
/
a(9,4,2)
/ \
b(9,2) c(4)
/
c(2)
Example2
Input:
<"a", 10>
<"c", 41>
<"b", 50>
<"abc", 5>
Output: <a[10,5]<b[5]<c[5]<>>>b[50]<>c[41]<>>
Solution & Analysis
@xfsm1912:
1.这题是字典树的变形题。其实就是建立字典树结构,同时在每个字符节点上不断向top10=[]这个list中添加给定的frequency。
2.遍历字典树的方法就是之前的传统方法,首先得到当前节点字典children中letter对应的TrieNode()作为child,如果没有letter在这个children里,就返回None,然后把child 定义为TrieNode(),在children中对应letter。然后执行addFrequency()。然后node = child往下移动。
3.在添加frequency时,如果新来的数比之前的数大,就不断交换,直到碰上比它更大的数位置。同时还看看top10长度是不是超过了10
4.时间复杂度是o(len(words))
5.注意,是child添加top10元素
LintCode Official
/**
* This reference program is provided by @jiuzhang.com
* Copyright is reserved. Please indicate the source for forwarding
*/
/**
* Definition of TrieNode:
* public class TrieNode {
* public NavigableMap<Character, TrieNode> children;
* public List<Integer> top10;
* public TrieNode() {
* children = new TreeMap<Character, TrieNode>();
* List<Integer> top10 = new ArrayList<Integer>();
* }
* }
*/
public class TrieService {
private TrieNode root = null;
public TrieService() {
root = new TrieNode();
}
public TrieNode getRoot() {
// Return root of trie root, and
// lintcode will print the tree struct.
return root;
}
// @param word a string
// @param frequency an integer
public void insert(String word, int frequency) {
// Write your cod here
TrieNode cur = root;
int n = word.length();
for (int i = 0; i < n; ++i) {
Character c = word.charAt(i);
if (!cur.children.containsKey(c))
cur.children.put(c, new TrieNode());
cur = cur.children.get(c);
addFrequency(cur.top10, frequency);
}
}
public void addFrequency(List<Integer> top10, int frequency) {
top10.add(frequency);
int n = top10.size();
int index = n - 1;
while (index > 0) {
if (top10.get(index) > top10.get(index - 1)) {
int temp1 = top10.get(index);
int temp2 = top10.get(index - 1);
top10.set(index, temp2);
top10.set(index - 1, temp1);
index -= 1;
} else
break;
}
if (n > 10)
top10.remove(n - 1);
}
}
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