阳光明媚的菲越

2022年4月19日

摘要： Stack solution: class Solution { public int scoreOfParentheses(String s) { Stack<Integer> st = new Stack<>(); int score = 0; for(int i = 0; i < s.leng 阅读全文

posted @ 2022-04-19 10:19 阳光明媚的菲越阅读(17) 评论(0) 推荐(0)

311. Sparse Matrix Multiplication

摘要： My first solution, time complexityO(m*n), where m is the cell number of mat1, and n is the cell number of mat2: class Solution { public int[][] multip 阅读全文

posted @ 2022-04-19 08:15 阳光明媚的菲越阅读(17) 评论(0) 推荐(0)

1539. Kth Missing Positive Number

摘要： My first binary search solution: class Solution { public int findKthPositive(int[] arr, int k) { int l=0, r = arr.length-1; while(l+1<r){ int mid = (l 阅读全文

posted @ 2022-04-19 06:53 阳光明媚的菲越阅读(16) 评论(0) 推荐(0)

2022年4月14日

1060. Missing Element in Sorted Array

摘要： My solution: class Solution { public int missingElement(int[] nums, int k) { int i=1; for(;i<nums.length;i++){ int diff = nums[i]-nums[i-1]-1; if(diff 阅读全文

posted @ 2022-04-14 11:43 阳光明媚的菲越阅读(18) 评论(0) 推荐(0)

348. Design Tic-Tac-Toe

摘要： class TicTacToe { int n; int[] rows; int[] cols; int diag=0; int antiDiag = 0; public TicTacToe(int n) { this.n = n; rows = new int[n]; cols = new int 阅读全文

posted @ 2022-04-14 05:29 阳光明媚的菲越阅读(20) 评论(0) 推荐(0)

863. All Nodes Distance K in Binary Tree

摘要： The key to solve this problem is to find the path from root to target, and put the length to target of every node from root to target to a map. Then e 阅读全文

posted @ 2022-04-14 05:11 阳光明媚的菲越阅读(18) 评论(0) 推荐(0)

270. Closest Binary Search Tree Value

摘要： PreOrder: class Solution { double min = Integer.MAX_VALUE; int val; public int closestValue(TreeNode root, double target) { preOrder(root, target); re 阅读全文

posted @ 2022-04-14 02:47 阳光明媚的菲越阅读(10) 评论(0) 推荐(0)

616. Add Bold Tag in String

摘要： This is a merge interval's variety: class Solution { public String addBoldTag(String s, String[] words) { List<int[]> intervals = new ArrayList<>(); f 阅读全文

posted @ 2022-04-14 02:29 阳光明媚的菲越阅读(25) 评论(0) 推荐(0)

2022年4月12日

536. Construct Binary Tree from String

摘要： My solution, using stack: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode( 阅读全文

posted @ 2022-04-12 12:34 阳光明媚的菲越阅读(15) 评论(0) 推荐(0)

778. Swim in Rising Water

摘要： BFS, using PriorityQueue to poll out the smallest number every time. The largest number you get will be the result. class Solution { public int swimIn 阅读全文

posted @ 2022-04-12 07:30 阳光明媚的菲越阅读(10) 评论(0) 推荐(0)