LeetCode Unique Paths II

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

题目：

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.

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: m and n will be at most 100.

题解：

是Unique Paths的进阶版题目。思路与Unique Paths相似, 加了障碍物.

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:初始化是点dp[0][0].

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

AC Java:

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
 4             return 0;
 5         }
 6         int m = obstacleGrid.length;
 7         int n = obstacleGrid[0].length;
 8         
 9         int [][] dp = new int[m][n];
10         dp[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
11         
12         for(int i = 1; i){
13             dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i-1][0];
14         }
15         
16         for(int j = 1; j){
17             dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j-1];
18         }
19         
20         for(int i = 1; i){
21             for(int j = 1; j){
22                 dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i-1][j] + dp[i][j-1];
23             }
24         }
25         return dp[m-1][n-1];
26     }
27 }

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:

 1 class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
 4             return 0;
 5         }
 6         
 7         int m = obstacleGrid.length;
 8         int n = obstacleGrid[0].length;
 9         
10         int [] dp = new int[n];
11         dp[0] = obstacleGrid[0][0] == 1 ? 0 : 1;
12         
13         for(int j = 1; j){
14             dp[j] = obstacleGrid[0][j] == 1 ? 0 : dp[j-1];
15         }
16         
17         for(int i = 1; i){
18             for(int j = 0; j){
19                 if(j == 0){
20                     dp[j] = obstacleGrid[i][j] == 1 ? 0 : dp[j];
21                 }else{
22                     dp[j] = obstacleGrid[i][j] == 1 ? 0 : dp[j] + dp[j-1];
23                 }
24             }
25         }
26         return dp[n-1];
27     }
28 }

 

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4824959.html