02 2018 档案
摘要：https://leetcode.com/problems/symmetric-tree/description/Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree [1,2,2,3,4,4,3] is ...
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摘要：https://leetcode.com/problems/same-tree/description/Given two binary trees, write a function to check if they are the same or not.Two binary trees are considered the same if they are structurally iden...
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摘要：https://leetcode.com/problems/count-univalue-subtrees/description/Given a binary tree, count the number of uni-value subtrees.A Uni-value subtree means all nodes of the subtree have the same value.For...
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摘要：网上还有一种做法用STACK 现在无法理解，以后再弄！
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摘要：given an array, and an element to insert, and the position to insert this element,return a new array with the element inserted1,2,3,4,5,6 -> 1,2,3,15,4,5,6 1 public static void main(String[] args) {...
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摘要：用一次遍历，一个指针完成 2,3,8,4,1 -> 1,4,8,3,2 1 public static void main(String[] args) { 2 int[] org = {2,3,8,4,1 } ; 3 int[] res = reverseArray(org); 4 print(res); 5 } 6 7 ...
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摘要：https://leetcode.com/problems/invert-binary-tree/description/Invert a binary tree. 4 / \ 2 7 / \ / \1 3 6 9to 4 / \ 7 2 / \ / \9 6 3 1time: o(n) : n nodes space:...
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摘要：https://leetcode.com/problems/binary-tree-inorder-traversal/description/ Given a binary tree, return the inorder traversal of its nodes' values. For e
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摘要：https://leetcode.com/problems/binary-tree-preorder-traversal/description/ Given a binary tree, return the preorder traversal of its nodes' values. For
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摘要：https://leetcode.com/problems/maximum-depth-of-binary-tree/description/Given a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node do...
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摘要：https://leetcode.com/problems/balanced-binary-tree/description/Given a binary tree, determine if it is height-balanced.For this problem, a height-balanced binary tree is defined as:a binary tree in wh...
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摘要：https://leetcode.com/problems/binary-tree-postorder-traversal/description/ Description Implement an iterative, post-order traversal of a given binary
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摘要：https://leetcode.com/problems/sort-list/description/Sort a linked list in O(n log n) time using constant space complexity.找到中点之后的处理，和 143 "Reorder List" 是一样的 : 然后把tail 传后面去 ListNode tail = mid.next ;/...
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摘要：https://leetcode.com/problems/intersection-of-two-arrays/description/Given two arrays, write a function to compute their intersection.Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].Not...
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摘要：https://leetcode.com/problems/reorder-list/description/Given a singly linked list L: L0→L1→…→Ln-1→Ln,reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…You must do this in-place without altering the nodes' values....
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摘要：取中点： 翻转： 合并: 插入：使用DUMMY head 避免了头部不确定， 头部为空等各种情况
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摘要：https://leetcode.com/problems/find-peak-element/description/A peak element is an element that is greater than its neighbors.Given an input array where num[i] ≠ num[i+1], find a peak element and return...
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摘要：这个笨方法也能通过
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摘要：与Find Minimum in Rotated Sorted Array类似。因为rotate, 所以不能直接用Binary Search, 需要进行 二次判定。 case 1: nums[mid] == target, return mid. case 2: nums[mid] < nums[r
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摘要：Given a 2D matrix that contains integers only, which each row is sorted in an ascending order. The first element of next row is larger than (or equal
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摘要：Given a target integer T and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to T. Assumptions There can
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摘要：Given a target integer T and an integer array A sorted in ascending order, find the index of the last occurrence of T in A or return -1 if there is no
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摘要：Given a target integer T and an integer array A sorted in ascending order, find the index of the first occurrence of T in A or return -1 if there is n
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摘要：Given a target integer T and an integer array A sorted in ascending order, find the index i such that A[i] == T or return -1 if there is no such index
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摘要：（1） 先找中点（2） 再reverse后半段（3） 不用管两个子linked list的长度是否相等，从各个子链表的头开始一个一个比较。值不相等就返回false，相等就继续移动。 当取中点时，一定要 fast.next.next != null
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摘要：设：链表头是X，环的第一个节点是Y，slow和fast第一次的交点是Z。各段的长度分别是a,b,c，如图所示。环的长度是L。slow和fast的速度分别是qs,qf。
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摘要：https://leetcode.com/problems/add-two-numbers/description/You are given two non-empty linked lists representing two non-negative integers.The digits are stored in reverse order and each of their nodes...
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摘要：time: o(n) complexity: o(1)
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摘要：230,82,83 是一类题 time o(1) space o(1)
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摘要：https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/Given a sorted linked list, delete all duplicates such that each element appear only once.For example,Given 1->1->2, return...
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摘要：https://leetcode.com/problems/partition-list/description/ Given a linked list and a target value T, partition it such that all nodes less than T are l
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摘要：Description Insert a value in a sorted linked list. Examples L = null, insert 1, return 1 -> null L = 1 -> 3 -> 5 -> null, insert 2, return 1 -> 2 ->
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摘要：https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/Given a linked list, remove the nth node from the end of list and return its head.For example, Given linked list: 1->2->3->...
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摘要：https://leetcode.com/problems/delete-node-in-a-linked-list/description/Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.Supposed the linked...
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摘要：千万注意，不要遍历两遍 什么 while(oddcurr) while(evencurr) 因为underline 的链表是一个，所以遍历第一遍之后就已经把 next 的关系搞混了， 后面那个遍历就不准确了。
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摘要：https://leetcode.com/problems/merge-two-sorted-lists/description/Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two ...
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摘要：2-1-4-3-5 5 自己就弹了，所以没有后面的处理，所以(head=3)head.next 还是 直接连 last pop newhead(5) 所以是 2-1-4-3-5 要深刻理解当时stack 里 head = 3 newhead=5 这一点！
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摘要：https://leetcode.com/problems/linked-list-cycle/description/ time: o(n) space: o(1) Follow up:Can you solve it without using extra space? if you use e
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摘要：recursive: assume Node2 is the head, subproblem return: null<-Node3 <- Node4 then you want to change to ListNode newHead = reverse(head.next) will gen
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摘要：time complexity: offer(), size(), isEmpty(): o(1) for poll and peek: worst case: offer n times, peek/poll only once: o(n) average case: offer n times,
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摘要：Stack: LIFO, the Stack is a deprecated interface we should use Deque interface. The implementation can use LinkedList or array push, pop, peek, size()
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摘要：queue: FIFO, Queue is interface, the implementation could be LinkedList or array offer, poll, peek, size() isEmpty()
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摘要：时间： o（n） 空间 o(1): 没有recursion, 只有当前一层的call stack sorting 如果是general sorting algo(1:不能对数组的性质加以限制） comparison based(2: sorting 只能通过两个数的比较来进行) ， 最好的时间复杂度
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摘要：时间复杂度， 空间复杂度 average: nlogn : logn 层 每层 n 次排序 n n/2 n/2 n/4 n/4 n/4 n/4 worst: 每次取最大或者最小值
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摘要：1 public class MergeSort { 2 3 public static void sortIntegers(int[] array) { 4 // write your code here 5 if (array == null || array.length ==0 ) { 6 return ; ...
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摘要：1 /* 2 * 1: time complexity o(n^2) 3 * 2: good performance for items around 10-20: better than merge sort and quick sort 4 * 3: no extra space needed 5 * */ 6 public class SelectSort { 7 p...
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