677 Map Sum Pairs

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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

这题我取巧了,用了startWith。时间 insert: O(1) sum: O(n * len(key)) 每次都要遍历所有key

	Map<String, Integer> map;

	public MapSum() {
		map = new HashMap<String, Integer>();
	}

	public void insert(String key, int val) {
		if (map.containsKey(key)) {
			map.remove(key);
		}
		map.put(key, val);
	}

	public int sum(String prefix) {
		int sum = 0;
		for (String s : map.keySet()) {
			if (s.startsWith(prefix)) {
				sum += map.get(s);
			}
		}
		return sum;
	}

看到一个时间是O(len(key))的方法,维护两个map,每次都更新比给定key短的所有key的value。 https://discuss.leetcode.com/topic/103908/simple-java-hashmap-solution-o-1-sum-and-o-len-key-insert

/** Initialize your data structure here. */
Map<String, Integer> map;
Map<String, Integer> original;
public MapSum() {
    map = new HashMap<>();
    original = new HashMap<>();
}

public void insert(String key, int val) {
    val -= original.getOrDefault(key, 0); // calculate the diff to be added to prefixes
    String s = "";
    for(char c : key.toCharArray()) {
        s += c; // creating all prefixes
        map.put(s, map.getOrDefault(s, 0) + val); //update/insert all prefixes with new value
    }
    original.put(key, original.getOrDefault(key, 0) + val);
}

public int sum(String prefix) {
    return map.getOrDefault(prefix, 0);
}