[LeetCode] 63. Unique Paths II

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?

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

不同路径II。题意跟版本一一样，唯一的不同点在于路径上会有障碍物。依然是返回到底有多少不同的路径能从左上角走到右下角。

思路还是跟版本一一样，做法也是一维DP，唯一的不同点在于遇到障碍物的时候，DP数组里面这个坐标的值要置为0，说明是没有办法从起点走到这个点的。所以res数组初始化的时候，如果res[0]设置成1，扫描的范围需要从i = 0, j = 0开始，而版本一这个地方是从i = 0, j = 1开始的。

时间O(mn)

空间O(n)

Java实现

 1 class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         int length = obstacleGrid[0].length;
 4         int[] res = new int[length];
 5         res[0] = 1;
 6         for (int i = 0; i < obstacleGrid.length; i++) {
 7             for (int j = 0; j < obstacleGrid[0].length; j++) {
 8                 if (obstacleGrid[i][j] == 1) {
 9                     res[j] = 0;
10                 } else if (j > 0) {
11                     res[j] += res[j - 1];
12                 }
13             }
14         }
15         return res[length - 1];
16     }
17 }

JavaScript实现

 1 /**
 2  * @param {number[][]} obstacleGrid
 3  * @return {number}
 4  */
 5 var uniquePathsWithObstacles = function (obstacleGrid) {
 6     let len = obstacleGrid[0].length;
 7     let res = new Array(len).fill(0);
 8     res[0] = 1;
 9     for (let i = 0; i < obstacleGrid.length; i++) {
10         for (let j = 0; j < obstacleGrid[0].length; j++) {
11             if (obstacleGrid[i][j] == 1) {
12                 res[j] = 0;
13             } else if (j > 0) {
14                 res[j] += res[j - 1];
15             }
16         }
17     }
18     return res[len - 1];
19 };