[数据结构与算法]LeetCode 432. 全 O(1) 的数据结构

题目链接：

力扣https://leetcode-cn.com/problems/all-oone-data-structure/

【分析】能力不够，只能用TreeMap来做了，TreeMap的键存元素的出现次数，值为一个Set存出现了这些次的字符串，再用一个HashMap来进行String到出现次数的映射，插入和删除的时候先去查HashMap，再修改TreeMap。

class AllOne {

    Map<String, Integer> map = new HashMap<>();
    TreeMap<Integer, Set<String>> freq = new TreeMap<>();

    public AllOne() {

    }

    public void show(){
        int i;
        for(Integer it: freq.keySet()){
            System.out.print(it);
            System.out.print(" ");
        }
        System.out.println("   *****");
    }

    public void inc(String key) {
        if(!map.containsKey(key)){
            map.put(key, 1);
            if(!freq.containsKey(1)){
                Set<String> set = new HashSet<>();
                set.add(key);
                freq.put(1, set);
            }else{
                freq.get(1).add(key);
            }
        }else{
            int val = map.get(key);
            map.put(key, val + 1);
            freq.get(val).remove(key);
            if(freq.get(val).size() == 0) freq.remove(val);
            if(!freq.containsKey(val + 1)){
                Set<String> set = new HashSet<>();
                set.add(key);
                freq.put(val + 1, set);
            }else{
                freq.get(val + 1).add(key);
            }
        }
        // show();
    }

    public void dec(String key) {
        int val = map.get(key);
        if(val == 1){
            map.remove(key);
            freq.get(val).remove(key);
            if(freq.get(val).size() == 0) freq.remove(val);
        }else{
            map.put(key, map.get(key) - 1);
            freq.get(val).remove(key);
            if(freq.get(val).size() == 0) freq.remove(val);
            if(!freq.containsKey(val - 1)){
                HashSet<String> set = new HashSet<>();
                set.add(key);
                freq.put(val - 1, set);
            }else{
                freq.get(val - 1).add(key);
            }
        }
        // show();
    }

    public String getMaxKey() {
        if(freq.size() == 0) return "";
        Set<String> set = freq.get(freq.lastKey());
        return (String)set.toArray()[0];
    }

    public String getMinKey() {
        if(freq.size() == 0) return "";
        Set<String> set = freq.get(freq.firstKey());
        return (String)set.toArray()[0];
    }
}

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