{ "data": { "question": { "questionId": "63", "questionFrontendId": "63", "categoryTitle": "Algorithms", "boundTopicId": 1514, "title": "Unique Paths II", "titleSlug": "unique-paths-ii", "content": "
You are given an m x n
integer array grid
. There is a robot initially located at the top-left corner (i.e., grid[0][0]
). The robot tries to move to the bottom-right corner (i.e., grid[m-1][n-1]
). The robot can only move either down or right at any point in time.
An obstacle and space are marked as 1
or 0
respectively in grid
. A path that the robot takes cannot include any square that is an obstacle.
Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
\n\nThe testcases are generated so that the answer will be less than or equal to 2 * 109
.
\n
Example 1:
\n\n\nInput: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]\nOutput: 2\nExplanation: There is one obstacle in the middle of the 3x3 grid above.\nThere are two ways to reach the bottom-right corner:\n1. Right -> Right -> Down -> Down\n2. Down -> Down -> Right -> Right\n\n\n
Example 2:
\n\n\nInput: obstacleGrid = [[0,1],[0,0]]\nOutput: 1\n\n\n
\n
Constraints:
\n\nm == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
obstacleGrid[i][j]
is 0
or 1
.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”)。
\n\n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?
\n\n网格中的障碍物和空位置分别用 1
和 0
来表示。
\n\n
示例 1:
\n\n\n输入:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]\n输出:2\n解释:3x3 网格的正中间有一个障碍物。\n从左上角到右下角一共有 2
条不同的路径:\n1. 向右 -> 向右 -> 向下 -> 向下\n2. 向下 -> 向下 -> 向右 -> 向右\n
\n\n示例 2:
\n\n\n输入:obstacleGrid = [[0,1],[0,0]]\n输出:1\n\n\n
\n\n
提示:
\n\nm == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
obstacleGrid[i][j]
为 0
或 1
\r\nif obstacleGrid[i][j] is not an obstacle\r\n obstacleGrid[i,j] = obstacleGrid[i,j - 1] \r\nelse\r\n obstacleGrid[i,j] = 0\r\n\r\n\r\nYou can do a similar processing for finding out the number of ways of reaching the cells in the first column.", "For any other cell, we can find out the number of ways of reaching it, by making use of the number of ways of reaching the cell directly above it and the cell to the left of it in the grid. This is because these are the only two directions from which the robot can come to the current cell.", "Since we are making use of pre-computed values along the iteration, this becomes a dynamic programming problem.\r\n\r\n
\r\nif obstacleGrid[i][j] is not an obstacle\r\n obstacleGrid[i,j] = obstacleGrid[i,j - 1] + obstacleGrid[i - 1][j]\r\nelse\r\n obstacleGrid[i,j] = 0\r\n\r\n\r\n" ], "solution": null, "status": null, "sampleTestCase": "[[0,0,0],[0,1,0],[0,0,0]]", "metaData": "{\r\n \"name\": \"uniquePathsWithObstacles\",\r\n \"params\": [\r\n {\r\n \"name\": \"obstacleGrid\",\r\n \"type\": \"integer[][]\"\r\n }\r\n ],\r\n \"return\": {\r\n \"type\": \"integer\"\r\n }\r\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"
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