63-Unique Paths II

题目

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.

分析

相比于Unique Paths II，处理的时候加一个判断

如果该处有障碍，则把pathNum 设为0

为了使得第一行和第一列处理方便，所以申请的数组为(rows+1)*(cols+1)

实现

/*

Author:Fancy
Date:2017-03-20
Algorithm:63-Unique Paths II
Time Complexity:
*/

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int rows = obstacleGrid.size(), cols = obstacleGrid[0].size();
vector<vector<int>> pathNum(rows + 1, vector<int>(cols + 1, 0));
for (int i = 1; i < rows + 1; i++)
for (int j = 1; j < cols + 1; j++)
{
if (i == 1 && j == 1 && obstacleGrid[i-1][j-1]==0)
pathNum[i][j] = 1;
else if (obstacleGrid[i-1][j-1] == 1)
pathNum[i][j] = 0;
else
{
pathNum[i][j] = pathNum[i - 1][j] + pathNum[i][j - 1];
}
}
return pathNum[rows][cols];

}
};