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]<>>
思路：先遍历一遍数，然后重构树进行求解
/**
* Definition of TrieNode:
* public class TrieNode {
* public NavigableMap<Character, TrieNode> children;
* public List<Integer> top10;
* public TrieNode() {
* children = new TreeMap<Character, TrieNode>();
* 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);
}
}