Unique Paths

题目地址：
https://leetcode.com/problems/unique-paths/#/description

题目：

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.

解题思路：
这道题主要就是dp思想。这道题有一个follow up：当有obstacle的时候，需要首先对该点进行判断，即使是当i或者j是0的时候，如果是obstacle，那么马上将dp设为0.

代码：

public int uniquePaths(int m, int n) {
    int[][] dp = new int[m][n];
    for(int i = 0; i <= m - 1; i++){
        for(int j = 0 ; j <= n - 1; j++){
            if(i == 0 || j == 0){
                dp[i][j] = 1;
            }
            else{
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
    }
    return dp[m - 1][n - 1];
}

follow up：

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
        return 0;
    }
    for(int i = 0; i < obstacleGrid.length; i++){
        for(int j = 0; j < obstacleGrid[0].length; j++){
            if(obstacleGrid[i][j] == 1){
                obstacleGrid[i][j] = 0;
            }
            else if(i == 0 && j == 0){
                obstacleGrid[i][j] = 1;
            }
            else if(i == 0){
                obstacleGrid[i][j] = obstacleGrid[i][j - 1];
            }
            else if(j == 0){
                obstacleGrid[i][j] = obstacleGrid[i - 1][j];
            }
            else{
                obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
            }
        }
    }
    return obstacleGrid[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
}