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
*
* 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
* 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
* 现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/unique-paths-ii
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*
* 网格中的障碍物和空位置分别用 1 和 0 来表示。
*
*
* 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) {
//这个题很有动态规划的倾向
//上一个位置到最后一个位置有几个走法
//a[i][j] = a[i - 1][j] + a[i][j -1] 分别只能向右向下
int res[][] = new int[obstacleGrid.length][obstacleGrid[0].length];
//初始化，如果遇到障碍，那么那个位置不可到达为0
//左边只有一种走法，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;
}
//进行动态规划
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
*
* 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
* 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
* 问总共有多少条不同的路径？
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/unique-paths
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*
* @Author xiaof
* @Date 2019/7/15 22:28
* @Version 1.0
**/
public class UniquePaths {
public int solution(int m, int n) {
//这个题很有动态规划的倾向
//上一个位置到最后一个位置有几个走法
//a[i][j] = a[i - 1][j] + a[i][j -1] 分别只能向右向下
int res[][] = new int[m][n];
//初始化，如果遇到障碍，那么那个位置不可到达为0
//左边只有一种走法，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;
}
//进行动态规划
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.
*
* 我们在两条独立的水平线上按给定的顺序写下 A 和 B 中的整数。
* 现在，我们可以绘制一些连接两个数字 A[i] 和 B[j] 的直线，只要 A[i] == B[j]，且我们绘制的直线不与任何其他连线（非水平线）相交。
* 以这种方法绘制线条，并返回我们可以绘制的最大连线数。
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/uncrossed-lines
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*
* @Author xiaof
* @Date 2019/7/15 22:34
* @Version 1.0
**/
public class MaxUncrossedLines {
public int solution(int[] A, int[] B) {
//不能相交也就是用过的数前面就不能进行连接
//每当A出i个数，B出j个数的时候，可以进行连接的情况是,当第i\和j的位置正好可以连线
//如果不能，那么就分别加上A的i个数，和机上B的j个的时候取最大的一遍
//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,默认为0
for(int i = 1; i <= m; ++i) {
//A使用几个数
for(int j = 1; j <= n; ++j) {
//这个循环代表B使用几个数
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];
}
}
原文地址：https://www.cnblogs.com/cutter-point/p/11192239.html
时间： 2024-10-10 00:21:09