AXIMI的博客

提交代码class Solution { public boolean canVisitAllRooms(List<List<Integer>> rooms) { boolean[] access=new boolean[rooms.size()]; Arrays.fill(access, false); access[0] = true; Queue<Integer> next=new Linke

创建两个集合用于保存两组节点首先0放在集合1，0的邻节点放在集合21的邻节点是0和2，0已经在集合1里面了，把2也放到集合1里面去3的邻节点是0和2，都在集合1里面。所有节点都放在了集合1和2中，返回true（如果存放过程中出现矛盾返回false）提交代码class Solution { public boolean isBipartite(int[][] graph) { Set<Integer> g1=new HashSet<>(); S.

提交代码class Solution { public int minDays(int[] bloomDay, int m, int k) { int[] bd=bloomDay.clone(); Arrays.sort(bloomDay); int i=0,j=bloomDay.length-1,mid=(i+j)/2; while(i<j) { if(i==j-1) { if(hasBouquet(

介绍Bellman-Ford 算法是一种用于计算带权有向图中单源最短路径（SSSP：Single-Source Shortest Path）的算法。该算法由 Richard Bellman 和 Lester Ford 分别发表于 1958 年和 1956 年，而实际上 Edward F. Moore 也在 1957 年发布了相同的算法，因此，此算法也常被称为 Bellman-Ford-Moore 算法。Bellman-Ford 算法和 Dijkstra 算法同为解决单源最短路径的算法。对于带权有向图 G

bellman-ford算法提交代码class Solution { public int networkDelayTime(int[][] times, int N, int K) { int[][] graph = new int[N+1][N+1]; for(int i=0;i<N+1;i++) Arrays.fill(graph[i], -1); for(int[] t: times) graph[

提交代码class Solution { Map<Integer, List<Integer>> graph=new HashMap<>(); Set<Integer> allNodes = new HashSet<>(); public int[] findRedundantConnection(int[][] edges) { for(int[] e: edges) { if(findPath(e[0], e

欧拉通路提交代码class Solution { Map<String, PriorityQueue<String>> graph =new HashMap<>(); List<String> path=new LinkedList<>(); public List<String> findItinerary(List<List<String>> tickets) { for(L

提交代码class Solution { public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) { String start, end; Map<String, Map<String, Double>> graph=new HashMap<>(

二分法提交代码class Solution { public int kthSmallest(int[][] matrix, int k) { int low = matrix[0][0], high = matrix[matrix.length-1][matrix[0].length-1]; while(low<high) { int mid = low+(high-low)/2; int curRank = 0, j=matrix[0].le

解决思路以这棵树为例，首先MHT的根节点不会是叶子节点，所以删除0，1，2，5剩下的3，4就是MHT的根节点。即一层一层的删掉叶子节点，直到只剩下一个或者两个节点。提交代码class Solution{ public List<Integer> findMinHeightTrees(int n, int[][] edges) { List<Integer> res= new ArrayList<>(); List<Node> tre

提交代码class Solution { public int firstUniqChar(String s) { HashMap<Character, Integer> map=new HashMap<>(); for(int i=0;i<s.length();i++) map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0)+1); for(int

用递归提交代码class Solution { public int minTime(int n, int[][] edges, List<Boolean> hasApple) { TreeNode[] nodes=new TreeNode[n]; for(int i=0;i<n;i++) nodes[i]=new TreeNode(i); for(int[] e:edges

提交代码class Solution { public int countLargestGroup(int n) { int digitSum, maxSize=0,res=0; Map<Integer, Integer> map=new HashMap<>(); for(int i=1;i<=n;i++) { ...

提交代码class Trie { ArrayList<String> trie; /** Initialize your data structure here. */ public Trie() { trie=new ArrayList<>(); } /** Inserts a word into the tr...

提交代码class Solution { public int longestSubstring(String s, int k) { if(s==null) return 0; int res=0, curMax=0; for(int i=0;i<=s.length()-k;i++) { curMax ...

提交代码class Solution { public int minNumberOfFrogs(String croakOfFrogs) { ArrayList<Integer> allCroaks=new ArrayList<Integer>(); char[] chars=croakOfFrogs.toCharArray(); ...

提交代码class MedianFinder { private ArrayList<Integer> nums; /** initialize your data structure here. */ public MedianFinder() { nums=new ArrayList<>(); } public...

思路我的想法是用一张表来记录出现的数字及对应出现的次数。然后对所有数字的出现次数排名。然后出现排名前k的数字就是目标数字。提交代码class Solution { public List<Integer> topKFrequent(int[] nums, int k) { HashMap<Integer, Integer> map=new H...

提交代码class Solution { public List<String> fizzBuzz(int n) { List<String> res=new ArrayList<>(); for(int i=1;i<=n;i++) { if(i%3==0&&i%5==0) res...

解题思路四个数组中，找出组合之和为0的个数。一般想到的是4个for循环，时间复杂度O(n^4)，耗时太久了。可以考虑牺牲空间换取时间。将A B两个数组中数字的组合值用map保存起来，然后再遍历C D数组。综合时间复杂度O(N^2).提交代码class Solution { public int fourSumCount(int[] A, int[] B, int[] C, int[]...

题目描述给出数字n，要求返回所有1-n能能构成的二叉搜索树的根节点数组。用递归当输入为0时，输出是空数组提交代码class Solution { public List<TreeNode> generateTrees(int n) { if(n==0) return new ArrayList<TreeNode>(); re...

解题思路其实不用DP，用最简单的滑动窗口即可。提交代码class Solution { public int maxSubArray(int[] nums) { int curSum=nums[0],res=nums[0]; for(int i=1,len=nums.length;i<len;i++) { if(curSum+nums...

解题思路这道题的意思是，检查一串数字是不是一棵树先序遍历的结果。做法是使用递归代码class Solution { public boolean isValidSerialization(String preorder) { if(preorder == null || preorder.length()==0) return true; Array...

解题思路设置两个链表头，一个串接单数节点，一个串接双数节点。最后将两个链表的数据连接起来。时间复杂度为O(n)，是原始链表的长度空间复杂度为O(1)，用于保存两个新创建的链表头。代码class Solution { public ListNode oddEvenList(ListNode head) { ListNode trueHead = new ListN...

解题思路这道题就是找规律，找到规律以后就比较简单提交代码class Solution {public boolean canWinNim(int n) { return n%4==0?false:true; }}运行结果

提交代码class Solution { public boolean wordPattern(String pattern, String str) { char[] chars = pattern.toCharArray(); String[] words = str.split(" "); if(chars.length!=words.length) ...

Description：Given a string, find the length of the longest substring without repeating characters.Example1:Input: “abcabcbb”Output: 3Explanation: The answer is “abc”, with the length of 3.Ex...

Description：There are two sorted arrays nums1 and nums2 of size m and n respectively.Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).You may assume...

Description：Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.Example1:Input: “babad”Output: “bab”Note: “aba” is also a valid a...

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 contain a single digit. Add the two number...

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Description：Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that eac...

Description：Implement atoi which converts a string to an integer.The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, star...

Description：The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H N...

Description：Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, ...

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 lists.Example:Input: 1-&gt;2-&gt;4, 1-&gt;3-&gt...

题目链接提交代码：int romanToInt(char* s) { int i,p = 0,numslen=0; int sum = 0; int* nums = (int*)malloc(30 * sizeof(int)); while (s[p] != '\0') { if (s[p] == 'M') nums[p] = 1000; else if (s...

Description：Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives ...

Description：Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.Example1:Input: 121Output: trueExample2:Input: -121Output: f...

[ 题目链接 ]提交代码（递归）：class Solution { List&lt;String&gt; generateParenthesis(int n) { List&lt;String&gt; list=new ArrayList&lt;&gt;(); append(n,n,"",list); return list;} void append(in...