[leetcode]Minimum Path Sum

动态规划。就是要注意第0行和第0列的初始化。

public class Solution {
    public int minPathSum(int[][] grid) {
        // Start typing your Java solution below
        // DO NOT write main() function        
        int m = grid.length;
        if (m == 0) return 0;
        int n = grid[0].length;
        if (n == 0) return 0;
        int[][] mx = new int[m][n];
        
        mx[0][0] = grid[0][0];
                
        for (int i = 1; i < m; i++) {
            mx[i][0] = mx[i-1][0] + grid[i][0];
        }
        
        for (int j = 1; j < n; j++) {
            mx[0][j] = mx[0][j-1] + grid[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                mx[i][j] = Math.min(mx[i-1][j], mx[i][j-1]) + grid[i][j];
            }
        }
        
        return mx[m-1][n-1];
    }   
}

Python3

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        if len(grid) == 0 or len(grid[0]) == 0:
            return 0
        
        m = len(grid)
        n = len(grid[0])
        memo = {}
        
        for i in range(m):
            for j in range(n):
                if i == 0 and j == 0:
                    memo[(i, j)] = grid[i][j]
                elif i == 0:
                    memo[(i, j)] = grid[i][j] + memo[(i, j - 1)]
                elif j == 0:
                    memo[(i, j)] = grid[i][j] + memo[(i - 1, j)]
                else: # i != 0 or j != 0
                    memo[(i, j)] = grid[i][j] + min(memo[(i - 1, j)], memo[(i, j - 1)])
                    
        return memo[(m - 1, n - 1)]