LeetCode-62. Unique Paths/63. Unique Paths II

62. Unique Paths

Problem:

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).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Analysis:
这个题就是简单的DP，经典实现方法有两种，一种递归，一种while循环dp，其中循环dp可以有更节省空间的方法。

Answer:
递归：

public class Solution {
    int[][] num ;
    public int uniquePaths(int m, int n) {
        num = new int[m][n];
        return dfs(0,0,m,n);
    }
    public int dfs(int i,int j,int m,int n){
        if(i==m-1 || j==n-1) return 1;
        if(num[i][j]>0) return num[i][j];
        num[i][j]=dfs(i+1,j,m,n)+dfs(i,j+1,m,n);
        return num[i][j];
    }
}

while循dp：

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for(int i=0;i<m;i++){
            dp[i][0] =1;
        }
        for(int i=0;i<n;i++){
            dp[0][i]=1;
        }

        for(int i=1;i<m;i++){
           for(int j=1;j<n;j++){
               dp[i][j]= dp[i-1][j]+dp[i][j-1];
           }
        }
        return dp[m-1][n-1];
    }
}

while循dp改进：

public class Solution {
    public int uniquePaths(int m, int n) {
        int[]dp = new int[n];
        for(int i=0;i<n;i++){
            dp[i] =1;
        }

        for(int i=1;i<m;i++){
           for(int j=1;j<n;j++){
               dp[j]= dp[j-1]+dp[j];
           }
        }
        return dp[n-1];
    }
}

63. Unique Paths II

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.

Analysis:

• 比上一题加了阻碍条件，所以要做阻碍处理。
• 注意第0行和第0列。如果没有障碍物，所有路径都是1，但是如果有障碍物。从第一个有障碍物开始，这一排和这一列以后的所有路径是0，就是以后的地方都到不了。
• 利用flag标记将第一行之后的dp标记为0，并在遍历的时候注意第一列如果有1，则将dp[0]=0。

Answer:

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if(obstacleGrid[m-1][n-1]==1) return 0;
        int[] dp = new int[n];
        int flag=0;//第一行为1的位置
        for(int i=0;i<n;i++){
            if(obstacleGrid[0][i]==0) dp[i]=1;
            else{
                dp[i]=0;
                flag=i;
                break;
            }
        }
        //使第一行从1开始后的dp都为0
        if(flag>0){
            for(int i=flag+1;i<n;i++){
                dp[i]=0;
            }
        }
        for(int i=1;i<m;i++){
            if(obstacleGrid[i][0]==1) dp[0]=0;//解决第一列有1的情况，主要防止类似{{0,0},{1,0}}的情况
            for(int j=1;j<n;j++){
                if(obstacleGrid[i][j]==1) dp[j]=0;//做判断区别
                else{
                    dp[j]=dp[j-1]+dp[j];
                }
            }
        }
        return dp[n-1];
    }
}