leetcode 677. Map Sum Pairs

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本文介绍了一种使用字典树（Trie）实现 MapSumPairs 的方法，该方法包含 insert 和 sum 两个核心操作。insert 方法用于插入键值对，如果键已存在则更新其值；sum 方法返回所有键以指定前缀开始的键值对的值之和。通过详细的代码解释了如何处理键的添加与覆盖，并展示了如何高效地完成 sum 操作。

677. Map Sum Pairs

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.

Example 1:

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

1、这个题的一个难点在于怎么判断是添加还是覆盖。

class TrieNode{    
public:    
    char var;    
    int value; 
    bool isword;
    TrieNode* children[26];    
        
    TrieNode()       
    {        
        var = 0;        
        value = 0;
        isword = 0;
        memset(children, 0, sizeof(TreeNode*)*26);        
        
    }        
    TrieNode(char c, int v)      
    {        
        var = c;        
        value = v; 
        isword = 0;
        memset(children, 0, sizeof(TreeNode*)*26);        
    }        
};  

class MapSum {
public:
    /** Initialize your data structure here. */
    MapSum() 
    {
        root = new TrieNode();    
    }
    
    void insert(string s, int val) 
    {
        TrieNode* p = root;    
        //先要判断是覆盖还是添加，也就是判断是否存在这个string
        int isexist = 0;
        for (int k = 0; k < s.size(); k++)    
        {    
            if (p->children[s[k] - 'a'] != 0) 
                p = p->children[s[k] - 'a'];
            else
                break;
            if (k + 1 == s.size() && p->isword) //说明已经存在
            {
                 isexist = p->value; //那就是覆盖 需要先减去原来的值
            }
        }
        //再改变每个节点的value,有公共前缀的，就把值都加上去。
        p = root;
        for (int k = 0; k < s.size(); k++)    
        {    
            if (p->children[s[k] - 'a'] == 0)    
            {    
                TrieNode* pnew = new TrieNode(s[k], val);    
                p->children[s[k] - 'a'] = pnew; 
                p = p->children[s[k] - 'a']; 
            }  
            else
            {
                p = p->children[s[k] - 'a'];
                p->value += val - isexist;  //如果是覆盖，减去原来的值
            }
            
            if (k + 1 == s.size())
                p->isword = true;
        }        
    }
    
    int sum(string s) 
    {
        TrieNode* p = root;    
        for (int k = 0; k < s.size(); k++)    
        {        
            if (p->children[s[k] - 'a'] != 0)
                p = p->children[s[k] - 'a']; 
            else
                return 0;
        }  
        return p->value;      
    }
private:
    TrieNode* root;
};

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