LeetCode 63. Unique Paths II

原题链接在这里：https://leetcode.com/problems/unique-paths-ii/

题目：

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

题解：

DP的更新当前点方式有所不同, 若是此处有障碍物. dp[i][j]里这一点就是0, 若此处没有障碍物, dp[i][j]是这一点就同行上一列和同列上一行的和.

Note: dp[0][0] initializatin as 0 when there is an obstacle there.

Time Complexity: O(m*n).

Space: O(m*n).

AC Java:

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
            return 0;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        
        int [][] dp = new int[m][n];
        dp[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
        
        for(int i = 1; i<m; i++){
            dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i-1][0];
        }
        
        for(int j = 1; j<n; j++){
            dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j-1];
        }
        
        for(int i = 1; i<m; i++){
            for(int j = 1; j<n; j++){
                dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

对应的本题也有像Unique Paths一般的降维存储历史信息的方法.

Note:更新时这里j是从0开始与Unique Paths不同，因为首列上有obstacle时需要更新dp[j] = dp[j]. 就是上一行的首列值.

Time Complexity: O(m*n). Space: O(n).

AC Java:

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
            return 0;
        }
        
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        
        int [] dp = new int[n];
        dp[0] = obstacleGrid[0][0] == 1 ? 0 : 1;
        
        for(int j = 1; j<n; j++){
            dp[j] = obstacleGrid[0][j] == 1 ? 0 : dp[j-1];
        }
        
        for(int i = 1; i<m; i++){
            for(int j = 0; j<n; j++){
                if(j == 0){
                    dp[j] = obstacleGrid[i][j] == 1 ? 0 : dp[j];
                }else{
                    dp[j] = obstacleGrid[i][j] == 1 ? 0 : dp[j] + dp[j-1];
                }
            }
        }
        return dp[n-1];
    }
}

类似Unique Paths, Minimum Path Sum.