62. Unique Paths

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 109.

代码：

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) continue;
                if (i - 1 >= 0) dp[i][j] += dp[i-1][j];
                if (j - 1 >= 0) dp[i][j] += dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

63. Unique Paths II

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

题目解析：

• 可以新建一个数组或者直接使用obstacleGrid数组来

代码：

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