比较地图并用相交的键更新一个，键的值最低

Question

我需要这个解决方案来解决 LeetCode 问题 1002：查找常用字符，但我没有找到合适的解决方案。

我需要的操作是用“a”和“b”之间的相交键更新映射“a”，并将键的值设置为两者中较低的值。

虽然这个用例很少，但有一个解决方案还是不错的。

class Solution {
    public List<String> commonChars(String[] A) {
        List<String> list = new ArrayList<>();
        Map<Character, Integer> a = new HashMap<>();
        Map<Character, Integer> b = new HashMap<>();
        List<Character> removeChars = new ArrayList<>();
        
        //base: populate map a so you can compare with next string in A[]
        
        for(char c : A[0].toCharArray()){
            int count = a.getOrDefault(c, 0);
            a.put(c, count+1);
        }
        
        /*
        compare each character in A[] to keys in map a
        
            if contains
                put character in map b and incremnt count
            else
                remove from map a
            
        update map a with whatever has smaller value for each key
        clear map b
            
        iterate
            
        put all keys in list for how many values it has
        
        */
        for(int i = 1; i < A.length; i++){
            for(char c : A[i].toCharArray()){
                if(a.containsKey(c)){
                    int count = b.getOrDefault(c, 0);
                    b.put(c, count+1);
                } else
                    a.remove(c);
            }


            /*
             Here I need to compare the keys to each map
                 - If they were present in both, take the lower value
                 - If not, then I needed to remove the key from map "a"
            */


            for(char c : removeChars){
                a.remove(c);
            }
            b.clear();
        }
        for(Map.Entry<Character, Integer> entry : a.entrySet()){
            char key = entry.getKey();
            int val = entry.getValue();
            for(int i = 0; i < val; i++){
                list.add(String.valueOf(key));
            }
        }
        return list;
        
    }
}

Answer 1

• 从地图 a 创建条目集
• 遍历条目
• 将键和值设置为变量以进行简单比较
  • 如果key之间有交集，用key的较低值覆盖map a
  • 否则将键添加到字符列表中

您不能在循环期间从条目集中删除键，而是将要删除的键存储在列表中，并在循环发生后将其删除。

Map<Character, Integer> a = new HashMap<>();
Map<Character, Integer> b = new HashMap<>();
List<Character> removeChars = new ArrayList<>();

for(Map.Entry<Character, Integer> entry1 : a.entrySet()){
    char keyA = entry1.getKey();
    int valA = entry1.getValue();
    if(b.containsKey(keyA)){
        int valB = b.get(keyA);
        a.put(keyA, Math.min(valA, valB));
     }else
        removeChars.add(keyA);
}

一般用途

Map<Key, Value> mapA = new HashMap<Key, Value>();
Map<Key, Value> mapB = new HashMap<Key, Value>();
List<T> remove = new ArrayList<T>();

for(Map.Entry<Ket, value> entry : mapA.entrySet()){
    Key key = entry.getKey();
    Value valA = entry.getValue();
    
    if(mapB.containskey(key)){
        int valB = mapB.get(key);
        mapA.put(key, Math.min(valA, valB);
    } else
        remove.add(keyA);
}