[LintCode/LeetCode] Unique Paths II

firim 发布于2019-08-14 15:29 / 2507人阅读

摘要：和完全一样的做法，只要在初始化首行和首列遇到时置零且即可。对了，数组其它元素遇到也要置零喏，不过就不要啦。

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];
    }
}

GPU云服务器云服务器 Paths unique Mysql_unique array_unique

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