62. Unique Paths 和63. Unique Paths II

另外64. Minimum Path Sum也和这两道题目是一类的题目，采用动态规划来解决，链接：http://blog.csdn.net/shanshanhi/article/details/56674313

一、62. 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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

方法一、数学方法：

int uniquePaths(int m, int n) {
        int N = m + n -;
        int k = m - ;
        double res = ;
        for(int i = ; i <= k; i++)
            res = res * (N - k + i) / i;
        return (int)res;
    }
           

方法二、动态规划的方法：

int uniquePaths(int m, int n) {
        vector<vector<int>> path(m,vector<int> (n,));
        for(int i = ; i < m; i++){
            for(int j = ; j < n; j++)
                path[i][j] = path[i-][j] + path[i][j-];
        }

        return path[m - ][n - ];
    }
           

二、63. Unique Paths II

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.

动态规划：

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

        return dp[m][n];
    }