[LC] 64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

class Solution {
    public int minPathSum(int[][] grid) {
        int row = grid.length;
        int col = grid[0].length;
        int[][] paths = new int[row][col];
        paths[0][0] = grid[0][0];
        // start from 1 should include the current grid value and prev value
        for (int i = 1; i < row; i++) {
            paths[i][0] = grid[i][0] + paths[i - 1][0];
        }
        for (int i = 1; i < col; i++) {
            paths[0][i] = grid[0][i] + paths[0][i - 1];
        }
        for (int i = 1; i < row; i++) {
            for (int j = 1; j < col; j++) {
                paths[i][j] = Math.min(paths[i - 1][j], paths[i][j - 1]) + grid[i][j];
            }
        }
        return paths[row - 1][col - 1];
    }
}

原文地址：https://www.cnblogs.com/xuanlu/p/12047225.html