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”.
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 il本文来源gao($daima.com搞@代@#码(网5lustrated 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.
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; j0) table[j] = table[j-1] + table[j]; } } return table[n-1]; }
转载请注明原文链接：CCI 9.2 机器人移动路径
