leetcode62

2019-11-06 08:20:49

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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? 感觉可以拉个小学生来做，这种题明显就是数学题qvq，既然只能向下向右走，那到达当前格子的路径数等于上面格子＋左边格子路径数，两层循环就得到了每个格子的路径数，也就得到最后一个格子路径数，显然最左边一列和最上面一行的格子路径数均为1，就可以用个二维容器描述，下面的定义没有区别只是为了方便嗯！

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.

增加了障碍带来的区别就是，当最左边一列和最上面一行出现障碍时，这个障碍格子后面或下面的格子都无法到达，也就是值将从1变为0，剩余的格子出现障碍，到达此格子的路径数变为0，对下一步格子路径数带来影响。

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;／／最左边一行 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]; }};

感觉代码仿佛可以变得简单一点，然鹅不想改嗯。。。

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