[Leetcode] Unique Paths 惟一路徑

Unique Paths I

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?

動態規劃

複雜度

時間 O(NM) 空間 O(NM)

思路

由於要走最短路徑，每一步只能向右方或者下方走。因此通過每個格子路徑數只可能源自左方或上方，這就獲得了動態規劃的遞推式，咱們用一個二維數組dp儲存每一個格子的路徑數，則dp[i][j]=dp[i-1][j]+dp[i][j-1]。最左邊和最上邊的路徑數都固定爲1，表明一直沿着最邊緣走的路徑。

代碼

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];
    }
}

一維數組

複雜度

時間 O(NM) 空間 O(N)

思路

實際上咱們能夠複用每一行的數組來節省空間，每一個元素更新前的值都是其在二維數組中對應列的上一行的值。這裏dp[i] = dp[i - 1] + dp[i];

代碼

public class Solution {
    public int uniquePaths(int m, int n) {
        int[] dp = new int[n];
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                // 每一行的第一個數都是1
                dp[j] = j == 0 ? 1 : dp[j - 1] + dp[j];
            }
        }
        return dp[n - 1];
    }
}

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.

動態規劃

複雜度

時間 O(NM) 空間 O(NM)

思路

解法和上題如出一轍，只是要判斷下當前格子是否是障礙，若是是障礙則通過的路徑數爲0。須要注意的是，最上面一行和最左邊一列，一旦遇到障礙就再也不賦1了，由於沿着邊走的那條路徑被封死了。

代碼

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

一維數組

複雜度

時間 O(NM) 空間 O(N)

思路

和I中的空間優化方法同樣，不過這裏判斷是不是障礙物，並且每行第一個點的值取決於上一行第一個點的值。

代碼

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