[LeetCode] 63. Unique Paths II

轉自LeetCode

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

<Solution>

Unique Path 的衍生題

這次增加的條件是，有些位置是無法到達的

那思路還是一樣，用 dynamic programming 的方式去解

解題想法如下

• 從位置 (0,0) 出發，如果遇到障礙物，就把該位置可到達步數清為0

• 如果沒遇到障礙物，就用 dp[i][j] = dp[i-1][j] + dp[i][j-1] 的概念去解

code 如下
c++

	class Solution {
	public:
	int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
	if(obstacleGrid.size() == 1 && obstacleGrid[0].size() == 1) {
	return (obstacleGrid[0][0] == 1) ? 0 : 1;
	}
	const int m = obstacleGrid.size(), n = obstacleGrid[0].size();
	vector<int> dp(n, 0);
	dp[0] = 1;
	for(int r = 0; r < m; r++) {
	for(int c = 0; c < n; c++) {
	if(obstacleGrid[r][c] == 1) {
	//> obstacle
	dp[c] = 0;
	}
	else if(c > 0) {
	dp[c] += dp[c-1];
	}
	}
	}
	return dp[n-1];
	}
	};

view raw findUniquePathsII.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun uniquePathsWithObstacles(obstacleGrid: Array<IntArray>): Int {
	return if(obstacleGrid[0][0] == 1) {
	0
	} else {
	val m = obstacleGrid.size
	val n = obstacleGrid[0].size
	val dp = Array(m) { IntArray(n) {0} }
	dp[0][0] = 1
	for(c in 1 until n) {
	dp[0][c] = if(obstacleGrid[0][c] == 1) 0 else dp[0][c-1]
	}
	for(r in 1 until m) {
	dp[r][0] = if(obstacleGrid[r][0] == 1) 0 else dp[r-1][0]
	}
	for(r in 1 until m) {
	for (c in 1 until n) {
	dp[r][c] = if(obstacleGrid[r][c] == 1) {
	0
	} else {
	dp[r-1][c] + dp[r][c-1]
	}
	}
	}
	dp.last().last()
	}
	}
	}

view raw unique_paths_II.kt hosted with ❤ by GitHub