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? 感覺可以拉個小學生來做，這種題明顯就是數(shù)學題qvq，既然只能向下向右走，那到達當前格子的路徑數(shù)等于上面格子＋左邊格子路徑數(shù)，兩層循環(huán)就得到了每個格子的路徑數(shù)，也就得到最后一個格子路徑數(shù)，顯然最左邊一列和最上面一行的格子路徑數(shù)均為1，就可以用個二維容器描述，下面的定義沒有區(qū)別只是為了方便嗯！
class Solution {public: int uniquePaths(int m, int n) { vector<vector<int>> result(m,vector<int>(n,1)); for(int i=1;i<m;i++) for(int j=1;j<n;j++){ result[i][j]=result[i-1][j]+result[i][j-1]; } return result[m-1][n-1]; }};2、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.
增加了障礙帶來的區(qū)別就是，當最左邊一列和最上面一行出現(xiàn)障礙時，這個障礙格子后面或下面的格子都無法到達，也就是值將從1變?yōu)?，剩余的格子出現(xiàn)障礙，到達此格子的路徑數(shù)變?yōu)?，對下一步格子路徑數(shù)帶來影響。
class Solution {public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m=obstacleGrid.size(); int n=obstacleGrid[0].size(); if(obstacleGrid[0][0]==1||obstacleGrid[m-1][n-1]==1) return 0; if(m==1&&n==1) return 1; vector<vector<int>> result(m,vector<int>(n,1)); for(int i=0;i<n-1;i++) if(obstacleGrid[0][i]==1) for(int j=i+1;j<=n-1;j++) result[0][j]=0; ／／最上面一行 for(int j=0;j<m-1;j++) if(obstacleGrid[j][0]==1) for(int i=j+1;i<=m-1;i++) result[i][0]=0;／／最左邊一行 for(int i=1;i<m;i++) for(int j=1;j<n;j++){ if(obstacleGrid[i-1][j]==1&&obstacleGrid[i][j-1]!=1) result[i][j]=result[i][j-1]; else if(obstacleGrid[i][j-1]==1&&obstacleGrid[i-1][j]!=1) result[i][j]=result[i-1][j]; else if(obstacleGrid[i-1][j]==0&&obstacleGrid[i][j-1]==0) result[i][j]=result[i-1][j]+result[i][j-1]; else result[i][j]=0; } return result[m-1][n-1]; }};感覺代碼仿佛可以變得簡單一點，然鵝不想改嗯。。。
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