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Question

Formatted question description: https://leetcode.ca/all/62.html

62	Unique Paths

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.

@tag-array

Algorithm

Using Dynamic Programming to solve, you can maintain a two-dimensional array dp, where dp[i][j] represents the number of different moves to the current position, and then the state transition equation can be obtained as: dp[i][j ] = dp[i-1][j] + dp[i][j-1], here in order to save space, use a one-dimensional array dp, refresh line by line.

Code

  • 
    public class Unique_Paths {
    
        class Solution {
            public int uniquePaths(int m, int n) {
    
                if (m <= 0 || n <= 0) {
                    return 0;
                }
    
                int[][] dp = new int[m][n];
    
                for (int i = 0; i < m; i++) {
                    for (int j = 0; j < n; j++) {
    
                        if (i == 0 || j == 0) {
                            dp[i][j] = 1;
                        } else { // i,j both not 1
                            dp[i][j] = dp[i-1][j] + dp[i][j-1];
                        }
                    }
                }
    
                return dp[m-1][n-1];
            }
        }
    
    
    }
    
    ############
    
    class Solution {
        public int uniquePaths(int m, int n) {
            int[][] dp = new int[m][n];
            for (int i = 0; i < m; ++i) {
                Arrays.fill(dp[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];
        }
    }
    
  • // OJ: https://leetcode.com/problems/unique-paths/
    // Time: O(MN)
    // Space: O(N)
    class Solution {
    public:
        int uniquePaths(int m, int n) {
            vector<int> dp(n + 1, 0);
            dp[n - 1] = 1;
            for (int i = m - 1; i >= 0; --i) {
                for (int j = n - 1; j >= 0; --j) dp[j] += dp[j + 1];
            }
            return dp[0];
        }
    };
    
  • class Solution:
        def uniquePaths(self, m: int, n: int) -> int:
            dp = [[1] * n for _ in range(m)]
            for i in range(1, m):
                for j in range(1, n):
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            return dp[-1][-1]
    
    ############
    
    class Solution(object):
      def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [1] * n
    
        for i in range(1, m):
          pre = 1
          for j in range(1, n):
            dp[j] = dp[j] + pre
            pre = dp[j]
        return dp[-1]
    
    
  • func uniquePaths(m int, n int) int {
    	dp := make([][]int, m)
    	for i := 0; i < m; i++ {
    		dp[i] = make([]int, n)
    	}
    	for i := 0; i < m; i++ {
    		for j := 0; j < n; j++ {
    			if i == 0 || j == 0 {
    				dp[i][j] = 1
    			} else {
    				dp[i][j] = dp[i-1][j] + dp[i][j-1]
    			}
    		}
    	}
    	return dp[m-1][n-1]
    }
    
  • function uniquePaths(m: number, n: number): number {
        let dp = Array.from({ length: m }, v => new Array(n).fill(1));
        for (let i = 1; i < m; ++i) {
            for (let j = 1; j < n; ++j) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
    
    
  • impl Solution {
        pub fn unique_paths(m: i32, n: i32) -> i32 {
            let (m, n) = (m as usize, n as usize);
            let mut dp = vec![1; n];
            for i in 1..m {
                for j in 1..n {
                    dp[j] += dp[j - 1];
                }
            }
            dp[n - 1]
        }
    }
    
    

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