63. Unique Paths II

Problem

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).

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.

Note: m and n will be at most 100.

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

Solution

/**
 * @param {number[][]} obstacleGrid
 * @return {number}
 */
var uniquePathsWithObstacles = function(obstacleGrid) {
  var m = obstacleGrid.length;
  var n = (obstacleGrid[0] || []).length;
  var dp = Array(m);
  var left = 0;
  var top = 0;

  if (!m || !n) return 0;

  for (var i = 0; i < m; i++) {
    dp[i] = Array(n);
    for (var j = 0; j < n; j++) {
      left = (j === 0 || obstacleGrid[i][j - 1] === 1) ? 0 : dp[i][j - 1];
      top = (i === 0 || obstacleGrid[i - 1][j] === 1) ? 0 : dp[i - 1][j];
      dp[i][j] = obstacleGrid[i][j] === 1 ? 0 : ((i === 0 && j === 0) ? 1 : (left + top));
    }
  }

  return dp[m - 1][n - 1];
};

Explain:

dp[i][j] 代表到达该点的路径数量。该点可以从左边点或上边点到达 也就是 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。

考虑特殊情况：

1. 该点为障碍，dp[i][j] = 0;
2. 左边点为障碍或不存在，left = 0;
3. 上边点点为障碍或不存在，top = 0;
4. 左边点与上边点均不存在，即起点，dp[i][j] = 1;

Complexity:

• Time complexity : O(m*n).
• Space complexity : O(m*n).