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leetcode-algorithms/README.md

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|500|[Keyboard Row](https://leetcode.com/problems/keyboard-row/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/KeyboardRow.java) | O(n) |O(1) | Easy|
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|494|[Target Sum](https://leetcode.com/problems/target-sum/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TargetSum.java) | O(2^n) |O(1) | Medium|
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|492|[Construct the Rectangle](https://leetcode.com/problems/construct-the-rectangle/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/ConstructTheRectangle.java) | O(n) |O(1) | Easy| Array
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|490|[The Maze](https://leetcode.com/problems/the-maze/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TheMaze.java) | O(m*n) |O(m*n) | Medium| BFS
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|485|[Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/MaxConsecutiveOnes.java) | O(n) |O(1) | Easy| Array
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|482|[License Key Formatting](https://leetcode.com/problems/license-key-formatting/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/LicenseKeyFormatting.java) | O(n) |O(n) | Medium|
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|477|[Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TotalHammingDistance.java) | O(n) |O(1) | Medium| Bit Manipulation
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package com.stevesun.solutions;
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import java.util.LinkedList;
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import java.util.Queue;
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/**
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* There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.
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Given the ball's start position, the destination and the maze, determine whether the ball could stop at the destination.
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The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.
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Example 1
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Input 1: a maze represented by a 2D array
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0 0 1 0 0
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0 0 0 0 0
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0 0 0 1 0
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1 1 0 1 1
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0 0 0 0 0
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Input 2: start coordinate (rowStart, colStart) = (0, 4)
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Input 3: destination coordinate (rowDest, colDest) = (4, 4)
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Output: true
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Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.
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Example 2
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Input 1: a maze represented by a 2D array
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0 0 1 0 0
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0 0 0 0 0
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0 0 0 1 0
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1 1 0 1 1
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0 0 0 0 0
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Input 2: start coordinate (rowStart, colStart) = (0, 4)
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Input 3: destination coordinate (rowDest, colDest) = (3, 2)
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Output: false
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Explanation: There is no way for the ball to stop at the destination.
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Note:
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There is only one ball and one destination in the maze.
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Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
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The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.
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The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.
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*/
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public class TheMaze {
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/**
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* BFS: the key part of this algorithm is that: this is a ball that won't stop rolling until it hits a wall.
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* This means we'll have to store all the empty spaces where the ball was forced to stop into the queue, these are
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* the only places where the next starting points could be.
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* Then we use BFS to traverse through all four directions of each starting point.
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* <p>
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* Also, another point to note is: it must be a stop point for the ball.
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*/
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public boolean hasPath(int[][] maze, int[] start, int[] destination) {
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if (start[0] == destination[0] && destination[0] == destination[1]) return true;
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final int[] directions = new int[]{-1, 0, 1, 0, -1};
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Queue<Point> queue = new LinkedList<>();
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queue.offer(new Point(start[0], start[1]));
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int m = maze.length, n = maze[0].length;
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boolean[][] visited = new boolean[m][n];
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visited[start[0]][start[1]] = true;
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while (!queue.isEmpty()) {
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Point curr = queue.poll();
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int x = curr.x, y = curr.y;//keep the original value
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for (int i = 0; i < directions.length - 1; i++) {
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int xx = x, yy = y;//use temp variables to move
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while (xx >= 0 && yy >= 0 && xx < m && yy < n && maze[xx][yy] == 0) {
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xx += directions[i];
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yy += directions[i + 1];
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}
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xx -= directions[i];
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yy -= directions[i + 1];
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if (visited[xx][yy]) continue;
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visited[xx][yy] = true;
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if (destination[0] == xx && destination[1] == yy) return true;
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queue.offer(new Point(xx, yy));
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}
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}
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return false;
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}
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class Point {
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int x;
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int y;
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public Point(int x, int y) {
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this.x = x;
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this.y = y;
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}
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}
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}
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package com.stevesun;
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import com.stevesun.solutions.TheMaze;
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import org.junit.Before;
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import org.junit.BeforeClass;
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import org.junit.Test;
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import static junit.framework.Assert.assertEquals;
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public class TheMazeTest {
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private static TheMaze test;
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private static boolean expected;
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private static boolean actual;
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private static int[][] maze;
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private static int[] start;
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private static int[] destination;
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@BeforeClass
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public static void setup(){
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test = new TheMaze();
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}
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@Before
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public void setupForEachTest(){}
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@Test
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public void test1(){
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maze = new int[][]{
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{0,0,0,0,0},
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{1,1,0,0,1},
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{0,0,0,0,0},
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{0,1,0,0,1},
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{0,1,0,0,0}
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};
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start = new int[]{4,3};
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destination = new int[]{0,1};
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actual = test.hasPath(maze, start, destination);
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expected = false;
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assertEquals(expected, actual);
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}
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@Test
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public void test2(){
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maze = new int[][]{
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{0,0,1,0,0},
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{0,0,0,0,0},
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{0,0,0,1,0},
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{1,1,0,1,1},
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{0,0,0,0,0}
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};
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start = new int[]{0,4};
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destination = new int[]{4,4};
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actual = test.hasPath(maze, start, destination);
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expected = true;
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assertEquals(expected, actual);
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}
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}

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