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一 题目

  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?

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

Note: m and n will be at most 100.

Example 1:

Input:[  [0,0,0],  [0,1,0],  [0,0,0]]Output: 2Explanation: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 -> Down2. Down -> Down -> Right -> Right

二 分析

   medium级别，这个题目跟上一个差不多，区别就是这个加了障碍物。

所以解题思路一样。区别是加了障碍物的判断。

DFS

public static void main(String[] args) {		// TODO Auto-generated method stub		int[][] obstacleGrid ={				{0,0,0},				{0,1,0},				{0,0,0}		};	int res = uniquePathsWithObstacles( obstacleGrid);	System.out.println(res);		}    //dfs	public static int uniquePathsWithObstacles(int[][] obstacleGrid) {		int m = obstacleGrid.length;		int n = obstacleGrid[0].length;		 if (obstacleGrid[0][0] != 0 ||	                obstacleGrid[m - 1][n - 1] != 0) return 0;		int[][] f = new int[m][n];		f[0][0] = obstacleGrid[0][0]==0?1:0;		return dfs(m-1,n-1,f,obstacleGrid);				    }		static int dfs(int x, int y, int[][] f,int[][] obstacleGrid) {        if (x < 0 || y < 0) return 0; // 数据非法，终止条件             if(obstacleGrid[x][y]==1){//障碍物        	return 0;        }        if (x == 0 && y == 0) return f[0][0]; // 回到起点，收敛条件                if (f[x][y] > 0) {            return f[x][y];        } else {            return f[x][y] = dfs(x - 1, y,f,obstacleGrid) +  dfs(x, y - 1,f,obstacleGrid);        }    }

Runtime: 4 ms, faster than 5.18% of Java online submissions for Unique Paths II.

Memory Usage: 40.9 MB, less than 15.38% of Java online submissions for Unique Paths II.

动态规划

  使用一个二维的 dp 数组，这里的 dp[i][j] 表示到达 (i, j) 位置的不同路径的数量，状态转移方程跟之前的一样：每个位置只能由其上面和左面的位置移动而来，也是由其上面和左边的 dp 值相加来更新当前的 dp 值，如下所示：

dp[i][j] = dp[i-1][j] + dp[i][j-1]

  另外，最上面的一行跟最左边的行初始化要判断下，不能全部为1，遇到障碍物就是0。

public static void main(String[] args) {		int[][] obstacleGrid ={				{0,0,0},				{0,1,0},				{0,0,0}		};	int res = uniquePathsWithObstacles( obstacleGrid);	System.out.println(res);		}		public static int uniquePathsWithObstacles(int[][] obstacleGrid) {		int m = obstacleGrid.length;		int n = obstacleGrid[0].length;				int[][] dp = new int[m][n];		//初始化第0列		for(int i=0;i

Runtime: 0 ms, faster than 100.00% of Java online submissions for Unique Paths II.

Memory Usage: 39.3 MB, less than 61.54% of Java online submissions for Unique Paths II.

时间复杂度O(M*N).

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