今天在做LeetCode的第194场周赛时的第二题时，使用List.contains方法来去重一直超时，在周赛快结束时才知道list的contains是从头开始进行比较，而Map.containsKey则是先通过Hash值得多对应的链表，在链表内进行比较，效率显然高于list。

题目

（1）先附上使用Map的方法

class Solution {
    public String[] getFolderNames(String[] names) {
        Map<String, Integer> differentNameCountMap = new HashMap();
        List<String> result = new ArrayList();
        for(String name : names) {
            Integer count = differentNameCountMap.get(name);
            if(count == null) {
                differentNameCountMap.put(name, 1);
                result.add(name);
                continue;
            }
            while(differentNameCountMap.containsKey(name + "(" + count + ")")) count ++;
            result.add(name + "(" + count + ")");
            differentNameCountMap.put(name, count + 1);
            differentNameCountMap.put(name + "(" + count + ")", 1);
        }
        String[] resultS = new String[result.size()];
        for(int i = 0; i < result.size(); i ++) resultS[i] = result.get(i);
        return resultS;
    }
}

（2）使用List的情况：

class Solution {
    public String[] getFolderNames(String[] names) {
        Map<String, Integer> differentNameCountMap = new HashMap();
        List<String> result = new ArrayList();
        for(String name : names) {
            //此处使用一次
            if(!result.contains(name)) {
                differentNameCountMap.put(name, 1);
                result.add(name);
                continue;
            }
            Integer count = differentNameCountMap.get(name);
            if(count == null) count = 1;
            
            //又使用一次
            while(result.contains(name + "(" + count + ")")) count ++;
            result.add(name + "(" + count + ")");
            differentNameCountMap.put(name, count + 1);
            differentNameCountMap.put(name + "(" + count + ")", 1);
        }
        String[] resultS = new String[result.size()];
        for(int i = 0; i < result.size(); i ++) resultS[i] = result.get(i);
        return resultS;
    }
}

源码分析：

1.ArrayList.contains()源码:

    /**
     * Returns <tt>true</tt> if this list contains the specified element.
     * More formally, returns <tt>true</tt> if and only if this list contains
     * at least one element <tt>e</tt> such that
     
     * @return <tt>true</tt> if this list contains the specified element
     */
    public boolean contains(Object o) {
        return indexOf(o) >= 0;
    }

    /**
     * Returns the index of the first occurrence of the specified element
     * in this list, or -1 if this list does not contain the element.
     */
    public int indexOf(Object o) {
        //不管o是否为空，数组都是从头开始进行比较！！
        
        if (o == null) {
            for (int i = 0; i < size; i++)
                if (elementData[i]==null)
                    return i;
        } else {
            for (int i = 0; i < size; i++)
                if (o.equals(elementData[i]))
                    return i;
        }
        return -1;
    }

2.Map.containsKey()的源码分析

    /**
     * Returns <tt>true</tt> if this map contains a mapping for the
     * specified key.
     *
     * @param   key   The key whose presence in this map is to be tested
     * @return <tt>true</tt> if this map contains a mapping for the specified
     * key.
     */
    public boolean containsKey(Object key) {
        return getNode(hash(key), key) != null;
    }
    
    /**
     * Implements Map.get and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @return the node, or null if none
     */
    final Node<K,V> getNode(int hash, Object key) {
        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {
            
            //进行hash值的比较
            //只有一个结点
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            
            //该hash值下有多个结点，判读是二叉树还是链表
            if ((e = first.next) != null) {
                //二叉树的比较方法
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

3.性能分析：

对比两个方法的源码可以得知，Map的containsKey()方法的查找性能明显优于List.contains()。 其实具体的查询效率，是Map是O(n1) 或O(log(n1))，而list则是O(n)

注：n1,表示当前hash值下的节点数，而n表示当前集合的所有元素的总数，明显Map的效率更高。

只能自己当时糊涂了啊！！！