63. Unique Paths II

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.

Note: m and n will be at most 100.

代码如下：

方案一：

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

方案二：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<int> dp(n+1,0);
        m--;
        int i = n-1;
        while(i >= 0 && obstacleGrid[m][i] != 1){
            dp[i] = 1;
            --i;
        }
        while(m-- > 0){ //这种循环判断可以将只有一行的情况统一考虑进去
            for(i = n-1; i >= 0; i--){
                dp[i] = (obstacleGrid[m][i] == 1)? 0 : dp[i+1] + dp[i];
            }
        }
        return dp[0];        
    }
};

解题思路：

• 用一个长度为n+1的vector对每一列dp ；
• 初始化一个长度为n+1，所有值都为0的dp；
• 最开始的时候初始化最后一行，如果不是障碍物，就将dp的对应位置变成1；
• 然后分别对倒数第二行，倒数第三行……第一行更新dp[i]，dp[i] = (obstacleGrid[m][i] == 1) ? 0 : dp[i] + dp[i+1];
• 最后dp中存储的是第一行的每一个非障碍物点到终点的路径数；
• dp[0]就是起点。