【LEETCODE】5六、數組分類，適中級別，題目：6二、6三、1035

package y2019.Algorithm.array.medium;

/**
 * @ClassName UniquePathsWithObstacles
 * @Description TODO 63. Unique Paths II
 *
 * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
 * The robot can only move either down or right at any point in time.
 * The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
 * Now consider if some obstacles are added to the grids. How many unique paths would there be?
 *
 * Input:
 * [
 *   [0,0,0],
 *   [0,1,0],
 *   [0,0,0]
 * ]
 * Output: 2
 * Explanation:
 * There is one obstacle in the middle of the 3x3 grid above.
 * There are two ways to reach the bottom-right corner:
 * 1. Right -> Right -> Down -> Down
 * 2. Down -> Down -> Right -> Right
 *
 * 一個機器人位於一個 m x n 網格的左上角 （起始點在下圖中標記爲「Start」 ）。
 * 機器人每次只能向下或者向右移動一步。機器人試圖達到網格的右下角（在下圖中標記爲「Finish」）。
 * 如今考慮網格中有障礙物。那麼從左上角到右下角將會有多少條不一樣的路徑？
 * 來源：力扣（LeetCode）
 * 連接：https://leetcode-cn.com/problems/unique-paths-ii
 * 著做權歸領釦網絡全部。商業轉載請聯繫官方受權，非商業轉載請註明出處。
 *
 * 網格中的障礙物和空位置分別用 1 和 0 來表示。
 *
 *
 * 1， 1， 1
 * 1， 0， 1
 * 1， 1， 2
 *
 * @Author xiaof
 * @Date 2019/7/15 22:00
 * @Version 1.0
 **/
public class UniquePathsWithObstacles {

    public int solution(int[][] obstacleGrid) {
        //這個題頗有動態規劃的傾向
        //上一個位置到最後一個位置有幾個走法
        //a[i][j] = a[i - 1][j] + a[i][j -1] 分別只能向右向下
        int res[][] = new int[obstacleGrid.length][obstacleGrid[0].length];

        //初始化，若是遇到障礙，那麼那個位置不可到達爲0
        //左邊只有一種走法，向下
        int h = 1,l = 1;
        for(int i = 0; i < obstacleGrid.length; ++i) {
            if(obstacleGrid[i][0] == 1) {
                h = 0;
            }
            res[i][0] =  h;
        }
        for(int j = 0; j < obstacleGrid[0].length; ++j) {
            if(obstacleGrid[0][j] == 1) {
                l = 0;
            }
            res[0][j] = l;
        }

        //進行動態規劃
        for(int i = 1; i < obstacleGrid.length; ++i) {
            for(int j = 1; j < obstacleGrid[i].length; ++j) {
                res[i][j] = obstacleGrid[i][j] == 1 ? 0 : res[i - 1][j] + res[i][j - 1];
            }
        }

        return res[obstacleGrid.length - 1][obstacleGrid[obstacleGrid.length - 1].length - 1];

    }

    public static void main(String[] args) {
        int data[][] = {{0,0,0},{0,1,0},{0,0,0}};
        UniquePathsWithObstacles fuc = new UniquePathsWithObstacles();
        System.out.println(fuc.solution(data));
        System.out.println(data);
    }

}

 

package y2019.Algorithm.array.medium;

/**
 * @ClassName UniquePaths
 * @Description TODO 62. Unique Paths
 * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
 * The robot can only move either down or right at any point in time.
 * The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
 * How many possible unique paths are there?
 *
 * Input: m = 3, n = 2
 * Output: 3
 * Explanation:
 * From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
 * 1. Right -> Right -> Down
 * 2. Right -> Down -> Right
 * 3. Down -> Right -> Right
 *
 * 一個機器人位於一個 m x n 網格的左上角 （起始點在下圖中標記爲「Start」 ）。
 * 機器人每次只能向下或者向右移動一步。機器人試圖達到網格的右下角（在下圖中標記爲「Finish」）。
 * 問總共有多少條不一樣的路徑？
 * 來源：力扣（LeetCode）
 * 連接：https://leetcode-cn.com/problems/unique-paths
 * 著做權歸領釦網絡全部。商業轉載請聯繫官方受權，非商業轉載請註明出處。
 *
 * @Author xiaof
 * @Date 2019/7/15 22:28
 * @Version 1.0
 **/
public class UniquePaths {

    public int solution(int m, int n) {
        //這個題頗有動態規劃的傾向
        //上一個位置到最後一個位置有幾個走法
        //a[i][j] = a[i - 1][j] + a[i][j -1] 分別只能向右向下
        int res[][] = new int[m][n];

        //初始化，若是遇到障礙，那麼那個位置不可到達爲0
        //左邊只有一種走法，向下
        int h = 1,l = 1;
        for(int i = 0; i < m; ++i) {
            res[i][0] =  h;
        }
        for(int j = 0; j < n; ++j) {
            res[0][j] = l;
        }

        //進行動態規劃
        for(int i = 1; i < m; ++i) {
            for(int j = 1; j < n; ++j) {
                res[i][j] = res[i - 1][j] + res[i][j - 1];
            }
        }

        return res[m - 1][n - 1];
    }
}

 

package y2019.Algorithm.array.medium;

/**
 * @ClassName MaxUncrossedLines
 * @Description TODO 1035. Uncrossed Lines
 *
 * We write the integers of A and B (in the order they are given) on two separate horizontal lines.
 * Now, we may draw connecting lines: a straight line connecting two numbers A[i] and B[j] such that:
 * A[i] == B[j];
 * The line we draw does not intersect any other connecting (non-horizontal) line.
 * Note that a connecting lines cannot intersect even at the endpoints: each number can only belong to one connecting line.
 * Return the maximum number of connecting lines we can draw in this way.
 *
 * Example 1:
 *
 * Input: A = [1,4,2], B = [1,2,4]
 * Output: 2
 * Explanation: We can draw 2 uncrossed lines as in the diagram.
 * We cannot draw 3 uncrossed lines, because the line from A[1]=4 to B[2]=4 will intersect the line from A[2]=2 to B[1]=2.
 *
 * 咱們在兩條獨立的水平線上按給定的順序寫下 A 和 B 中的整數。
 * 如今，咱們能夠繪製一些鏈接兩個數字 A[i] 和 B[j] 的直線，只要 A[i] == B[j]，且咱們繪製的直線不與任何其餘連線（非水平線）相交。
 * 以這種方法繪製線條，並返回咱們能夠繪製的最大連線數。
 * 來源：力扣（LeetCode）
 * 連接：https://leetcode-cn.com/problems/uncrossed-lines
 * 著做權歸領釦網絡全部。商業轉載請聯繫官方受權，非商業轉載請註明出處。
 *
 * @Author xiaof
 * @Date 2019/7/15 22:34
 * @Version 1.0
 **/
public class MaxUncrossedLines {

    public int solution(int[] A, int[] B) {
        //不能相交也就是用過的數前面就不能進行鏈接
        //每當A出i個數，B出j個數的時候，能夠進行鏈接的狀況是,當第i\和j的位置正好能夠連線
        //若是不能，那麼就分別加上A的i個數，和機上B的j個的時候取最大的一遍
        //res[i][j] = max{res[i - 1][j], res[i][j - 1]} or res[i][j] = res[i - 1][j - 1] + 1
        int m = A.length, n = B.length, dp[][] = new int[m + 1][n + 1];

        //初始化,默認爲0
        for(int i = 1; i <= m; ++i) {
            //A使用幾個數
            for(int j = 1; j <= n; ++j) {
                //這個循環表明B使用幾個數
                if(A[i - 1] == B[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[m][n];
    }

}