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This repository was archived by the owner on Apr 27, 2025. It is now read-only.

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Create 63. Unique Paths II.md
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‎63. Unique Paths II.md‎

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# 63. Unique Paths II
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### 2020年07月30日
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A robot is located at the top-left corner of a *m* x *n* grid (marked 'Start' in the diagram below).
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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).
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Now consider if some obstacles are added to the grids. How many unique paths would there be?
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![img](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
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An obstacle and empty space is marked as `1` and `0` respectively in the grid.
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**Note:** *m* and *n* will be at most 100.
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**Example 1:**
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```
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Input:
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[
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[0,0,0],
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[0,1,0],
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[0,0,0]
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]
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Output: 2
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Explanation:
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There is one obstacle in the middle of the 3x3 grid above.
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There are two ways to reach the bottom-right corner:
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1. Right -> Right -> Down -> Down
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2. Down -> Down -> Right -> Right
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```
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# Solution
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```swift
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class Solution {
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func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
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let m = obstacleGrid.count
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guard m > 0 else {
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return 0
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}
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let n = obstacleGrid[0].count
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guard n > 0 else {
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return 0
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}
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var cache = [[Int]].init(repeating: [Int].init(repeating: 0, count: n), count: m)
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for y in 1...m {
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for x in 1...n {
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if obstacleGrid[y - 1][x - 1] == 1 {
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cache[y - 1][x - 1] = 0
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} else if x == 1 {
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cache[y - 1][x - 1] = 1
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for _y in 1...y {
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if obstacleGrid[_y - 1][x - 1] == 1 {
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cache[y - 1][x - 1] = 0
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break
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}
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}
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} else if y == 1 {
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cache[y - 1][x - 1] = 1
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for _x in 1...x {
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if obstacleGrid[y - 1][_x - 1] == 1 {
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cache[y - 1][x - 1] = 0
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break
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}
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}
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} else {
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cache[y - 1][x - 1] = cache[y - 2][x - 1] + cache[y - 1][x - 2]
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}
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}
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}
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return cache[m - 1][n - 1]
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}
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}
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```
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