标签：spec color ons space finish tom bre top code
62. 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 the grid (marked ‘Finish‘ in the diagram below).
How many possible unique paths are there?
DP 注意边界
public class Solution { public int uniquePaths(int m, int n) { int dp[][] = new int[n][m]; for(int i = 0 ; i < n ; i++){ dp[i][0] = 1; } for(int j = 0 ; j < m ; j ++){ dp[0][j] = 1; } for(int i = 1; i < n ; i++){ for(int j = 1 ; j < m ; j++){ dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[n-1][m-1]; } }
63. 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 3x3 grid as illustrated below.
[ [0,0,0], [0,1,0], [0,0,0] ]
The total number of unique paths is 2.
public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1) return 0; int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length] ; for(int i = 0 ; i < obstacleGrid.length; i++){ if(obstacleGrid[i][0] != 1) dp[i][0] = 1; else break; } for(int i = 0 ; i < obstacleGrid[0].length; i++){ if(obstacleGrid[0][i] != 1) dp[0][i] = 1; else break; } for(int i = 1; i < obstacleGrid.length ; i++){ for(int j = 1; j < obstacleGrid[0].length ; j++){ if(obstacleGrid[i][j] == 1) dp[i][j] = 0; if(obstacleGrid[i][j] == 0) dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[obstacleGrid.length-1][obstacleGrid[0].length -1]; } }
标签：spec color ons space finish tom bre top code
原文地址：http://www.cnblogs.com/joannacode/p/6121156.html
