[LintCode/LeetCode] Unique Paths II-优快云博客

本文介绍了一种解决网格中存在障碍物时寻找从起点到终点的唯一路径数量的方法。文章详细阐述了两种动态规划算法：二维DP和一维DP，并提供了具体的实现代码。

Problem

Follow up for "Unique Paths":

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.

Notice

m and n will be at most 100.

Example

For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note

和Unique Path完全一样的做法，只要在初始化首行和首列遇到obstacle时置零且break即可。对了，数组其它元素遇到obstacle也要置零喏，不过就不要break啦。

Solution

二维DP

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

一维DP

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