随笔分类 - leetcode
题解
摘要：class Solution { public int bulbSwitch(int n) { return (int)Math.sqrt(n); // 刚开始灯都是关的，所以按奇数次会打开。比如n = 12； // 会在 第 1，12 2，6 3 ，4 轮被按，所以会被关闭 // 所以序号的因子个
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摘要：class Solution { private TrieNode root = new TrieNode(); public int minimumLengthEncoding(String[] words) { int res = 0; Arrays.sort(words,(o1,o2)->o2
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摘要：class Solution { public String removeDuplicateLetters(String s) { int n = s.length(); if(n <= 1) return s; Stack<Character> stack = new Stack<>(); for
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摘要：class Solution { int[] pa; int[] pb; public int find(int[] p, int x) { if(p[x] != x) p[x] = find(p,p[x]); return p[x]; } public int maxNumEdgesToRemov
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摘要：方法一：动态规划 class Solution { public: int minCost(string s, vector<int>& cost) { int n = s.size(),INF = 1e9; vector<vector<int>> f(n,vector<int>(27,INF));
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摘要：class Solution { int[] p; int[] s; public int largestIsland(int[][] grid) { if(grid.length == 0) return 0; int n = grid.length, m = grid[0].length; p
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摘要：方法一：树状数组 class NumArray { int[] nums; int[] bitArr; int n; public NumArray(int[] nums) { n = nums.length; this.nums = nums; bitArr = new int[n+1]; for
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摘要：class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { List<Integer> res = new ArrayList<>(); if (n == 1) { res.add(0); retu
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摘要：class Solution { public void moveZeroes(int[] nums) { int n = nums.length; for(int i = 0, j = 0; i < n; i++) { if(nums[i] != 0) { int t = nums[i]; num
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摘要：class Solution { public: vector<int> diffWaysToCompute(string input) { vector<int> res; int n = input.size(); for(int i = 0; i < n; i++) { char c = in
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摘要：方法一：记忆化dfs class Solution { boolean[][] visited; public boolean containsCycle(char[][] grid) { int m = grid.length, n = grid[0].length; visited = new
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摘要：分析： 方法一：贪心，首先将每个数的位置用map保存，遍历偶数索引的数，异或1找到它的情侣，如果不在当前索引i+1的位置上，就用map找到这个数，交换到i+1的位置上，并且更新i+1位置上的数在map中的位置 class Solution { public int minSwapsCouples(i
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摘要：class Solution { LinkedList<String> res = new LinkedList<>(); Map<String,PriorityQueue<String>> map = new HashMap<>(); public List<String> findItinera
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摘要：class Solution { public boolean judgePoint24(int[] nums) { double[] a = new double[]{nums[0],nums[1],nums[2],nums[3]}; return find(a); } public boolea
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摘要：class Solution { List<Integer> temp = new ArrayList<Integer>(); List<List<Integer>> ans = new ArrayList<List<Integer>>(); public List<List<Integer>> f
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摘要：分析：十叉树的遍历 class Solution { public List<Integer> lexicalOrder(int n) { List<Integer> res = new ArrayList<>(); for(int i = 1; i < 10; i++) { dfs(n,res,i
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摘要：分析： class Solution { public int subarrayBitwiseORs(int[] A) { Set<Integer> set = new HashSet<>(); Set<Integer> cur = new HashSet<>(); for(int num : A)
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摘要：class Solution { public int getLengthOfOptimalCompression(String s, int k) { int n = s.length(); int[][] dp = new int[n+1][k+1]; // dp[i][j]:考虑前i个字符最多
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摘要：class Solution { public int findNumberOfLIS(int[] nums) { int n = nums.length; int[] dp = new int[n]; Arrays.fill(dp,1); // dp[i] 以i结尾的最大长度 int[] coun
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摘要：class Solution { public List<Integer> largestDivisibleSubset(int[] nums) { List<Integer> res = new ArrayList<>(); int n = nums.length; if(n == 0) retu
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