LeetCode 677. Map Sum Pairs

一、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.

Example 1:

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

二、解题思路

在Hashmap中每个键值对应一个数，
插入一个键值对时：
如果key不存在，则将键值对存进Hashmap中，并且将key的所有前缀连同数值组成新的键值对存进Hashmap mem中；对于一个前缀，如果在mem中已经存在，那么只需在原数值上加上当前键值对的数值，如果前缀原先不存在于mem中，则直接赋为当前键值对的数值；
如果key存在，则key的新值会取代旧值，因此要在mem中更新与key有关的前缀的值

计算前缀对应的数值：
由于mem维护所有前缀对应的数值并在插入过程中一直保持更新，所以对只需取出mem中对应前缀的数值

三、Java代码

class MapSum {    Map<String,Integer> mem;    Map<String,Integer> set;    /** Initialize your data structure here. */    public MapSum() {        mem = new HashMap<>();        set = new HashMap<>();    }    public void insert(String key, int val) {        if (!set.containsKey(key)) {            set.put(key,val);            for (int i = 0; i <= key.length(); i++) {                String sub = key.substring(0,i);                mem.put(sub, mem.getOrDefault(sub,0) + val);            }        } else {            int cnt = set.get(key);            for (int i = 0; i <= key.length(); i++) {                String sub = key.substring(0,i);                mem.put(sub,mem.get(sub) - cnt + val);            }        }    }    public int sum(String prefix) {        if (!mem.containsKey(prefix)) return 0;        return mem.get(prefix);    }};/** * Your MapSum object will be instantiated and called as such: * MapSum obj = new MapSum(); * obj.insert(key,val); * int param_2 = obj.sum(prefix); */