总结一些比较好的学习资料

Java practice - http://www.javapractices.com/home/HomeAction.do
Openfire REST API version - https://www.igniterealtime.org/projects/openfire/plugins/1.3.7/restAPI/readme.html
TCP Retrans - https://accedian.com/enterprises/blog/network-packet-loss-retransmissions-and-duplicate-acknowledgements/

关于LDAP协议

LDAP 在网络七层结构中属于哪一层？
LDAP - 轻量级目录访问协议。属于会话层（Session Layer），在建立和管理session的时候使用。

怎样找到某个 domain 的 ldap server？

	luchens-MacBook-Pro:~ luchen$ nslookup -type=srv _ldap._tcp.google.com
	Server: 2001:568:ff09:10a::57
	Address: 2001:568:ff09:10a::57#53
	Non-authoritative answer:
	_ldap._tcp.google.com service = 5 0 389 ldap.google.com.

view raw ldap_nslookup.sh hosted with ❤ by GitHub

[LeetCode] 49. Group Anagrams

	class Solution {
	public List<List<String>> groupAnagrams(String[] strs) {
	List<List<String>> result = new ArrayList<>();
	Map<Character, Integer> characterMap;
	Map<Map<Character, Integer>, List<String>> groupMap = new HashMap<>();
	if (strs == null || strs.length == 0) {
	return result;
	}
	for (String s : strs) {
	characterMap = new HashMap<>();
	for (Character c : s.toCharArray()) {
	if (characterMap.get(c) == null) {
	characterMap.put(c, 1);
	} else {
	characterMap.put(c, characterMap.get(c) + 1);
	}
	}
	if (groupMap.get(characterMap) == null) {
	groupMap.put(characterMap, new ArrayList<>());
	}
	groupMap.get(characterMap).add(s);
	}
	for (List<String> group : groupMap.values()) {
	result.add(group);
	}
	return result;
	}
	}

view raw groupAnagrams.java hosted with ❤ by GitHub

[LeetCode] 18. 4Sum

\(▔▽▔)/

	class Solution {
	public List<List<Integer>> fourSum(int[] nums, int target) {
	List<List<Integer>> result = new ArrayList<>();
	Set<List<Integer>> resultSet = new HashSet<>();
	if (nums == null || nums.length < 4) {
	return result;
	}
	Arrays.sort(nums);
	for (int i = 0; i < nums.length - 3; i++) {
	for (int j = i + 1; j < nums.length - 2; j++) {
	if (j > i + 1 && nums[j] == nums[j - 1]) {
	continue;
	}
	int left = j + 1;
	int right = nums.length - 1;
	while (left < right) {
	int sum = nums[i] + nums[j] + nums[left] + nums[right];
	if (sum == target) {
	resultSet.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
	left++;// Attn!
	right--;
	} else if (sum > target) {
	right--;
	} else {
	left++;
	}
	}
	}
	}
	for (List<Integer> solution : resultSet) {
	result.add(solution);
	}
	return result;
	}
	}

view raw fourSum.java hosted with ❤ by GitHub

[LeetCode] 60. Permutation Sequence

有点想不明白，为什么要做 k = k - 1 ┑(￣▽ ￣)┍

	class Solution {
	public String getPermutation(int n, int k) {
	String chooseFrom = "123456789";
	int[] factorial = new int[n + 1];
	factorial[0] = 1;
	factorial[1] = 1;
	for (int i = 2; i <= n; i++) {
	factorial[i] = factorial[i - 1] * i;
	}
	k = k - 1; // Why?
	StringBuilder sb = new StringBuilder();
	for (int i = n; i >= 1; i--) {
	int x = k / factorial[i - 1];
	k = k % factorial[i - 1];
	sb.append(chooseFrom.charAt(x));
	chooseFrom = chooseFrom.substring(0, x) + chooseFrom.substring(x + 1, chooseFrom.length());
	}
	return sb.toString();
	}
	}

view raw getPermutation.java hosted with ❤ by GitHub

[LeetCode] 312. Burst Balloons

好难啊，不会做\(▔▽▔)/

	class Solution {
	public int maxCoins(int[] nums) {
	// dp[i][j] - max coins could get when burst all the balloons between i to j
	// dp[i][j] = max(dp[i][j], dp[i][k - 1] + nums[i - 1] * nums[k] * nums[j + 1] + dp[k + 1][j]), i <= k <= j
	if (nums == null || nums.length == 0) {
	return 0;
	}
	int[][] dp = new int[nums.length][nums.length];
	for (int len = 1; len <= nums.length; len++) {
	for (int i = 0; i + len <= nums.length; i++) {
	int j = i + len - 1;
	for (int k = i; k <= j; k++) {
	int left = i == 0 ? 1 : nums[i - 1];
	int right = j == nums.length - 1 ? 1 : nums[j + 1];
	int current = left * nums[k] * right;
	if (k - 1 >= i) {
	current = current + dp[i][k - 1];
	}
	if (k + 1 <= j) {
	current = current + dp[k + 1][j];
	}
	dp[i][j] = Math.max(dp[i][j], current);
	}
	}
	}
	return dp[0][nums.length - 1];
	}
	}

view raw maxCoins.java hosted with ❤ by GitHub

[LeetCode] 58. Length of Last Word

lol，工作很繁重，脑子不是很清楚，只能拿一道简单到逆天的题目充数了 ╮(￣▽￣)╭　

	class Solution {
	public int lengthOfLastWord(String s) {
	int end = s.length() - 1;
	while (end >= 0 && s.charAt(end) == ' ') {
	end--;
	}
	int start = end;
	while (start >= 0 && s.charAt(start) != ' ') {
	start--;
	}
	if (end >= 0) {
	return end - start;
	}
	return 0;
	}
	}

view raw lengthOfLastWord.java hosted with ❤ by GitHub