HashMap 拥有的方法：
#Easy Word Pattern
Given a pattern and a string str, find if str follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-emptyword in str.
Examples:
pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
pattern = "abba", str = "dog dog dog dog" should return false.
思路：因为要做到pattern str的一一对应，pattern的char 和str中的String一一对应，但是hashmap是个一对多的表，所以需要两个HashMap
首先需要判断char数组和String数组长度是否相等
#Hard Word Pattern II
Given a pattern and a string str, find if str follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-emptysubstring in str.
Examples:
pattern = "abab", str = "redblueredblue" should return true.
pattern = "aaaa", str = "asdasdasdasd" should return true.
pattern = "aabb", str = "xyzabcxzyabc" should return false.
思路：
必须要用递归才能遍历到所有的词，而且因为要确保一一对应，仍旧需要我们用两个表，这里可以用HashMap和HashSet,而不一定要用两个HashMap，
如果说这个char已经在map中有对应，那就看看是不是String的i这个位置以这个词开头，如果不是就false, 如果不在map中对应，就要String开始遍历可能的对应，每次记得要map和set一起加，不成功就map和set一起再减掉
#Medium 49.Group Anagrams
Given an array of strings, group anagrams together.
For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:
[
["ate", "eat","tea"],
["nat","tan"],
["bat"]
]
HashMap, We use the s sorted as the key, if containsKey so add s; if not, map .putkey and result.add. At last, Collections.sort()
#Medium 525.Contiguous Array
Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.
Example 1:
Input:[0,1]Output:2Explanation:[0, 1] is the longest contiguous subarray with equal number of 0 and 1.
Example 2:
Input:[0,1,0]Output:2Explanation:[0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
思路：使用HashMap, 如果不存在这个difference放入map, 如果存在就得到一个0，1的最大连续如果得到现在nums[i-1] ==0 就-1，如果不是就+1；difference[i] = difference[i - 1] + (nums[i-1] == 0 ? -1 : 1);
#Hard 76. Minimum Window Substring
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".
思路：使用一个map（数组）来记录pattern需要的字母的个数，使用一个map来记录现在我们看到的字母了，count来记录总共的需要字母个数。一旦count==0我们让start指针往后移，minLen来记录
#HARD 30. Substring with Concatenation of All Words
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9].
思路：用两个hashmap来实现，第一个hashmap是存我们的words中需要的字符串个数，然后利用for循环遍历，每次存下一个HashMap来看目前是不是已经满足了要求。
错误：map.containsKey
map.getOrDefault 用于得到目前的个数默认为0，然后再加1
