Map Sum Pairs问题及解法

问题描述：

Implement a MapSum class with 

insert

, and 

sum

 methods.

For the method 

insert

, you'll be given a pair of (string, integer). The string represents
the key and the integer represents the value. If the key already existed, then the original key-value pair will be overridden to the new one.

For the method 

sum

, you'll be given a string representing the prefix, and you need to
return the sum of all the pairs' value whose key starts with the prefix.

示例:

Input: insert("apple", 3), Output: Null
Input: sum("ap"), Output: 3
Input: insert("app", 2), Output: Null
Input: sum("ap"), Output: 5

问题分析：

对于含有公共前缀的字符串问题，我们可以考虑利用Trie数据结构，将字符串存储起来，每次求和时，若给定的公共前缀不存在，直接返回0；若存在，可以递归遍历出所有含有公共前缀的字串对应值的和。

过程详见代码：

class Trie{
public:
Trie * children[26];
bool isLeaf;
int val;
public:
Trie()
{
for (int i = 0; i < 26; i++) children[i] = nullptr;
isLeaf = false;
val = 0;
}
};

class MapSum {
public:
Trie * root;
/** Initialize your data structure here. */
MapSum() {
root = new Trie();
}

void insert(string key, int val) {
if (key == "") return;
Trie * tRoot = root;
for (int i = 0; i < key.length(); i++)
{
if (tRoot->children[key[i] - 'a'] == nullptr)
{
tRoot->children[key[i] - 'a'] = new Trie();
}
tRoot = tRoot->children[key[i] - 'a'];
}
tRoot->isLeaf = true;
tRoot->val = val;
}

int sum(string prefix) {
Trie * tRoot = root;
int res = 0;
for (int i = 0; i < prefix.length(); i++)
{
if (tRoot->children[prefix[i] - 'a'] != nullptr)
tRoot = tRoot->children[prefix[i] - 'a'];
else return 0;
}
blsum(tRoot, res);
return res;
}

void blsum(Trie * tRoot, int& res)
{
if (tRoot == nullptr) return;
if (tRoot->isLeaf) res += tRoot->val;

for (int i = 0; i < 26; i++)
{
blsum(tRoot->children[i], res);
}
}
};

/**
* Your MapSum object will be instantiated and called as such:
* MapSum obj = new MapSum();
* obj.insert(key,val);
* int param_2 = obj.sum(prefix);
*/