随笔分类 - sf
摘要：public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> result = new LinkedList<>(); if (root == null) { return result; } //
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摘要：一、朋友圈评论系统设计 A: {B, C, D} B: {A, D} 假设：A发表了一篇朋友圈文章，好友可以来评论 问题：表结构如何设计，评论插入和查询如何实现？ 查询需要有主题：图片或文字； 1 先获取所有评论 2 显示评论时，获取好友关心决定是否进行展示。 评论表： user_id,topic
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摘要：public static List<Integer> getVals(Node root){ List<Integer> ret=new ArrayList(); if(root == null) return ret; Queue<Node> qRet=new LinkedList(); qRe
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摘要：public static int sqrt(int a) { if (a < 0 || a == 0 || a == 1) return a; int startNum = 0; int endNum = a; for (; startNum + 1 < endNum; ) {//保证整数情况下，
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摘要：public void DFS(TreeNode root) {深度优先 Stack<TreeNode> stack = new Stack<>(); stack.add(root); while (!stack.isEmpty()) { // 移除最后一个 TreeNode tempNode =
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摘要：public static void main(String[] args) { int[] m = {2, 4, 6, 9, 12, 13, 15, 16}; int[] n = {3, 5, 9, 12, 15}; Search(m, n); } private static void Sear
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摘要：public Map<String, List<String>> getWordList(List<String> list) { Map<String, List<String>> ret = new HashMap<>(); for (int i = 0; i < list.size(); i+
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摘要：public static void main(String[] args) { Node n6 = new Node(6); Node n5 = new Node(5); n5.next = n6; Node n4 = new Node(4); n4.next = n5; Node n3 = ne
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摘要：public static boolean sum2(int ary[], int sum) { int start = 0; int end = ary.length - 1; for (; start < end; ) { if (ary[start] + ary[end] == sum) {
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摘要：public static void main(String[] args) { int ret = commpareStr("abc", "12345cab"); System.out.println(ret); } //orgStr源字段 public static int commpareSt
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摘要：核心思想找一个轴点元素，备份轴点元素，先从右往左找，end--操作。一旦不满足大小条件，则将右侧的数值放置到开始位置，然后开始位置进行++，向右侧找进行循环往复 public static void main(String[] args) { int ary[] = {6, 8, 8, 6, 2,
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摘要：public static void main(String[] args) { AtomicInteger a = new AtomicInteger(0); CountDownLatch countDownLatch = new CountDownLatch(1); Semaphore sema
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摘要：import com.google.common.base.Charsets; import com.google.common.base.Joiner; import com.google.common.base.Predicate; import com.google.common.base.S
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摘要：public ListNode getKthFromEnd(ListNode head, int k) { ListNode slow=head; ListNode fast=head; int t = 0; for(;fast!=null;){ if(t>=k) { slow=slow.next;
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摘要：private boolean slowAndFastSolution(ListNode head) { if (head == null) { return false; } ListNode slow = head; ListNode fast = head.next; while (fast
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摘要：public int add(int a, int b) { while (a != 0) { int temp = a ^ b; a = (a & b) << 1; b = temp; } return b; }
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摘要：public int divide(int dividend, int divisor) { boolean sign = (dividend > 0) ^ (divisor > 0); int result = 0; if(dividend>0) { dividend = -dividend; }
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摘要：public int divide(int dividend, int divisor) { //判断符号 int symbol=(dividend>0&&divisor>0)||(dividend<0&&divisor<0)?1:-1; //边界情况直接返回 if(dividend==Intege
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摘要：public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode first = dummy; ListNode second
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摘要：public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null) { return l2; } if(l2 == null) { return l1; } if(l1.val < l2.val) { l1.next =
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