【leetcode】63. Unique Paths II

一、题目描述

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.

思路：动态规划，还是用的“Unique Paths” 这道题的思想，不过设置了障碍之后，对于第一行和第一列要单独拿出来处理。用path[i][j] 表示到[i][j]位置的路径的条数，先判断[i][j]位置是否是障碍，即值是否是1，如果是障碍，那么path[i][j]=0，否则用状态转换方程path[i][j] = path[i-1][j] + path[i][j-1]

c++代码（3ms，26.70%）

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m,n;         m = obstacleGrid.size();        n = obstacleGrid[0].size();        int i,j;        int path[m][n];        path[0][0] = obstacleGrid[0][0]==0?1:0;        for(i=1; i<n;i++){  //先处理第一行            if(obstacleGrid[0][i] == 1){                path[0][i]=0;            }else{                path[0][i] = path[0][i-1];            }        }        for(i=1;i<m;i++){  //处理第一列            if(obstacleGrid[i][0] == 1){                path[i][0] = 0;            }else{                path[i][0] = path[i-1][0];            }                    }                for(i=1;i<m;i++){            for(j=1;j<n;j++){                if(obstacleGrid[i][j] == 1){                    path[i][j]=0;                }else{                    path[i][j] = path[i-1][j] + path[i][j-1];                }            }        }                return path[m-1][n-1];    }};

c++代码2（3ms）

往外扩展一行一列，左上角位置从1,1开始。这样对于第一行第一列就不需要单独拿出来讨论了。

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m,n;        m = obstacleGrid.size();        n = obstacleGrid[0].size();        int i,j;        vector<vector<int> > path(m+1, vector<int>(n+1, 0));        path[0][1] = 1;        for(i=1;i<=m;i++){            for(j=1;j<=n;j++){                if(!obstacleGrid[i-1][j-1]){                    path[i][j] = path[i-1][j] + path[i][j-1];                }            }        }//for        return path[m][n];    }};