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该问题来自 LeetCode 的第64题。
64. Minimum Path Sum原题链接：
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 1:
[[1,3,1], [1,5,1], [4,2,1]] Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.问题分析
该问题和另外一篇文章《 JAVA代码—算法基础：矩阵的最小路径和问题》描述的其实是一类问题。链接： 。都可以采用动态规划（Dynamic Programming）算法求解。
public int minPathSum(int[][] grid) { int m = grid.length;// row int n = grid[0].length; // column for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (i == 0 && j != 0) { grid[i][j] = grid[i][j] + grid[i][j - 1]; } else if (i != 0 && j == 0) { grid[i][j] = grid[i][j] + grid[i - 1][j]; } else if (i == 0 && j == 0) { grid[i][j] = grid[i][j]; } else { grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j]) + grid[i][j]; } } } return grid[m - 1][n - 1];} 简化一下算法：
public int minPathSum(int[][] grid) { int m = grid.length;// row int n = grid[0].length; // column for(int j = 1; j < n; j++) grid[0][j] = grid[0][j] + grid[0][j-1]; for(int i = 1; i < m; i++) grid[i][0] = grid[i][0] + grid[i-1][0]; for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j]) + grid[i][j]; } } return grid[m - 1][n - 1];} （完）
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