CCI 9.2 机器人移动路径

摄像有个机器人坐在X*Y网的左上角，只能想右、向下移动。机器人从(0,0)到(X,Y)有多少种走法？ 进阶 假设有些点为“禁区”，机器人不能踏足。设计一种算法，找出一条路径，让机器人从左上角移动到右下角。 这道题跟LeetCode上的Unique Paths 和Unique Paths I

摄像有个机器人坐在x*y网格的左上角，只能想右、向下移动。机器人从(0,0)到(x,y)有多少种走法？

进阶

假设有些点为“禁区”，机器人不能踏足。设计一种算法，找出一条路径，让机器人从左上角移动到右下角。

这道题跟LeetCode上的Unique Paths 和Unique Paths II一样。

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 thr grid(marked 'Finish' in the diagram below).

How many possible unique paths are there?

NOTE: m and n will be at most 100.

Unique Paths II

Follow up for "Unique Paths".

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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 3*3 grid as illustrated below.

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

NOTE: m and n will be at most 100.

解法：

Unique Paths

public int uniquePaths(int m, int n) {
        //这里用了DP解法，因为这种解法可以最大程度避免整数越界问题
        int[][] memo = new int[m][n];
        for(int i=0; i<m; i++)
            memo[i][0] = 1;
        for(int i=0; i<n; i++)
            memo[0][i] = 1;
        
        for(int i=1; i<m; i++)
            for(int j=1; j<n; j++)
                memo[i][j] = memo[i-1][j] + memo[i][j-1];
        
        return memo[m-1][n-1];
    }

Unique Paths II

这里用了一维数组来代替二维数组

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        if(m == 0) return 0;
        int n = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1) return 0;
        int[] table = new int[n];
        table[0] = 1;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(obstacleGrid[i][j] == 1)
                    table[j] = 0;
                else if(j>0)
                    table[j] = table[j-1] + table[j];
            }
        }
        return table[n-1];
    }