63. Unique Paths II

[Tcdian]

63. Unique Paths II

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

网格中的障碍物和空位置分别用 1 和 0 来表示。

Example 1

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Note

• 1 <= m, n <= 100

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[][]} obstacleGrid
 * @return {number}
 */
var uniquePathsWithObstacles = function(obstacleGrid) {
    const dp = Array.from(
        new Array(obstacleGrid.length),
        () => new Array(obstacleGrid[0].length).fill(0)
    );
    for (let i = 0; i < obstacleGrid.length; i++) {
        for (let j = 0; j < obstacleGrid[0].length; j++) {
            if (obstacleGrid[i][j] === 0) {
                if (i === 0 && j === 0) {
                    dp[i][j] = 1;
                } else if(i === 0) {
                    dp[i][j] = dp[i][j - 1];
                } else if (j === 0) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
                }
            }
        }
    }
    return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
};

• TypeScript Solution

function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    const dp: number[][] = Array.from(
        new Array(obstacleGrid.length),
        () => new Array(obstacleGrid[0].length).fill(0)
    );
    for (let i = 0; i < obstacleGrid.length; i++) {
        for (let j = 0; j < obstacleGrid[0].length; j++) {
            if (obstacleGrid[i][j] === 0) {
                if (i === 0 && j === 0) {
                    dp[i][j] = 1;
                } else if(i === 0) {
                    dp[i][j] = dp[i][j - 1];
                } else if (j === 0) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
                }
            }
        }
    }
    return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
};