HashMap原理及put方法與get方法的調用過程

/**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

/**
     * Implements Map.put and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

/**
     * Replaces all linked nodes in bin at index for given hash unless
     * table is too small, in which case resizes instead.
     */
    final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        // 如果數組等於null 或 數組長度小於 64
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
        // 重新散列，使得鏈表變短
            resize();
        // 如果hash沖突，且數組長度大於 64，則隻能使用紅黑樹結構
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
             // 返回新的紅黑樹
            do {
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

public class HashTest {
 public static void main(String[] args) {
  // 對"負載因子的大小對程序的影響規律"進行測試
  // threshold=capacity * loadFactor ---- 閾值 = 容量 x 負載因子
  // 源代碼擴容後容量是擴容前的二倍
  int n1 = 10;   // 對照組
  int n2 = 1000000; // put/get多少組
  long t0 = 0;      //總耗時
  float lf = 0.9f;  //負載因子
  int capacity = 100; //初始容量
  HashMap map = null;
  
  //對照組循環
  for (int j = 1; j <= n1; j++) {
   map = new HashMap(capacity, lf);
   List<String> list = new ArrayList<String>();
   // 利用循環進行put
   for (int i = 0; i < n2; i++) {
    String temp = HashTest.randomString();
    map.put(temp, i);
    list.add(temp);
   }
   long time = 0; // 總耗費時間
   // 利用循環get
   for (int i = 0; i < n2; i++) {
    String temp = list.get(i);
    long t1 = System.currentTimeMillis();
    map.get(temp);
    long t2 = System.currentTimeMillis();
    long t3 = t2 - t1;// 花費時間
    time += t3;
   }
   System.out.println("組"+j+"花費時間(ms)=" + time);
   t0 += time;
   map = null;
  }
  System.out.println("get出 "+n2+" 對鍵值對中，"+n1+"組數據得出:");
  System.out.println("---------------------------------");
  System.out.println("每get"+n2+"對鍵值對 平均花費時間(毫秒)："+(t0/n1));
 }
 /**
  * 產生隨機字符串方法 
  * @return
  */
 public static String randomString() {
  // 最終產生的字符串
  StringBuffer sb = new StringBuffer();
  // 字符串樣本
  String str = "回到傢卡薩恒大帝景阿薩德節快樂就看見瞭困窘企業無辜的鄙視你別這麼想按一個預告的哈上東國際按時大大伽伽匯頂科技啊啥看的撒打算大的歐亞報出去qwertyuiopasdfghjklzxvcbnm,.;p[']/\1234567890zxcvbnmaksjhfgdlpoiuytrewq阿斯加德克拉斯近段時間的書上方法更符合輔導費的冠福股份極樂空間流口水";
  // System.out.println("樣本字符串長度:"+str.length());
  // 產生一個1到30的數字
  int num = (int) (Math.random() * 30 + 1);
  // System.out.println("num="+num);
  // 用for循環從樣本字符串中提取出字符進行組合
  for (int i = 0; i < num; i++) {
   int num1 = (int) (Math.random() * str.length()); // 產生一個0到字符串樣本的數字
   // 根據索引值獲取對應的字符
   char charAt = str.charAt(num1);
   sb.append(charAt);
  }
  // System.out.println("產生一個長度為"+num+"的字符串");
  return sb.toString();
 }