Hiraccc_的博客

class Solution { public static int shortestPathBinaryMatrix(int[][] grid) { int [][] dir = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}}; int minLength[][] = new int[grid.length][grid.length]; Queue<int[]> queue.

用队列实现BFS: java.util.Queue初始化一个无穷大的数INF，准备一个距离二维数组，数组规模与输入相匹配，每一元素初始化为INF。这个距离二维数组可以记录从起点到当前位置的距离，也可以作为记忆化使用(即当前位置存储的距离 != INF)。迷宫地图：/*10#S######.#......#..#.#.##.##.#.#........##.##.####....#....#.#######.#....#......####.###.....#...G#*/

import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int n = scan.nextInt(); char[][] map = new char[n][n]; boolean[][] vis = new boolean[n][n];.

最后一个过不了干啥啥不行，写烂代码第一名我就是记录一下，建议别看import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int n = scan.nextInt(); char[][] map = new char[n][n];

class Solution { private int n = 0; public int findCircleNum(int[][] M) { n = M.length; int res = 0; boolean[] vis = new boolean[n]; for (int i = 0; i < n; i++) { if (vis[i] == false) { .

class Solution { private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; public List<List<Integer>> pacificAtlantic(int[][] grid) { List<List<Integer>> list = new ArrayList<>(); int[][] recLU.

class Solution { public int numIslands(char[][] grid) { int res = 0; for(int i = 0; i< grid.length;i++){ for(int j = 0; j<grid[0].length;j++){ if(dfs(grid,i,j)){ res+=1; .

class Solution { public int maxAreaOfIsland(int[][] grid) { int res = 0; int xLen = grid.length; int yLen = grid[0].length; for (int i = 0; i < xLen; i++) { for (int j = 0; j < yLen; j++) { .

import java.util.*;public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); char[] arrC = sc.next().toCharArray(); Stack<Character> sk = new Stack<>(); for (int i.

有一说一，这题30分，要考虑的点还是很多的。下面的写法有点麻烦。import java.util.Arrays;import java.util.Scanner;import java.util.TreeSet;public class Main { public static void main(String[] args) { Scanner scan = new Scanner(System.in); String s = scan.next(

原题数据本来是char数组误写为String数组了，将就看吧。注释的是输入输出功能。public class Main { public static void main(String[] args) {// Scanner scan = new Scanner(System.in);// int N = scan.nextInt();// int M = scan.nextInt(); int N = 10;

/*10 12W . . . . . . . . W W .. W W W . . . . . W W W. . . . W W . . . W W .. . . . . . . . . W W .. . . . . . . . . W . .. . W . . . . . . W . .. W . W . . . . . W W .W . W . W . . . . . W .. W . W . . . . . . W .. . W . . . . . . . W .*/im.

public class Main { public static void main(String[] args) { String s = "12345"; char[] c = s.toCharArray(); for (int i = 0; i < c.length; i++) { char [] cc = new char[i+1]; for (int j = 0; j < cc..

需要注意是恢复（回溯回去）。public class Main { public static void main(String[] args) { String s = "ABC"; permutation(0,s.length()-1,s.toCharArray()); } private static void permutation(int left, int right, char[] c) { if(left ==

用途：复杂度优化。public class Main { public static void main(String[] args) { System.out.println(fib(20)); } static int fib(int n){ int [] res = new int[25]; if(n==1 | n==2){ res[n] = 1; return res[n

先排序，然后两两数之间相减得到这个结果的次数最多的就是公差。然后再用（数组最大值-最小值）/公差 + 1 就是项数。import java.util.ArrayList;import java.util.Scanner;import java.util.TreeSet;public class Main { public static void main(String[] args) { Scanner scan = new Scanner(System.in)

本题的关键点在于注意完全二叉树并非满二叉树。可以根据深度强转整数进行判断：if(Math.log(n+1)/Math.log(2) > (int)(Math.log(n+1)/Math.log(2))){}完整代码：import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner scan = new Scanner(System

做法是按位取余，这里换个写法。转换为StringBuilder：public class Main { public static void main(String[] args) { int sum = 0; Scanner scan = new Scanner(System.in); int n = scan.nextInt(); for (int i = 1; i <= n; i++) { ..

times爆了 填空题public class Main { public static void main(String[] args) { int count = 0; for (int i = 1; i <= 1999; i++) { StringBuilder si = new StringBuilder(String.valueOf(i)); if(si.indexOf("2") == -1 &am

卡拉兹(Callatz)猜想：是对任何一个正整数n，如果它是偶数，那么把它砍掉一半；如果它是奇数，那么把(3n+1)砍掉一半。这样一直反复砍下去，最后一定在某一步得到n=1。题目描述：我们今天的题目不是证明卡拉兹猜想，而是对给定的任一不超过 1000 的正整数n，简单地数一下，需要多少步（砍几下）才能得到n=1？代码一：#include<stdio.h>int main(){ int n=0,i=0; scanf("%d",&n); whil...