2023.10.11

         本题与不同路径 思路类似，设定二维dp数组，dp[i][j]的含义为：从起点到坐标(i,j)的最小路径和。然后就是常规的动态规划做法了，直接看代码：

class Solution {public int minPathSum(int[][] grid) {int row = grid.length;int col = grid[0].length;int[][] dp = new int[row][col];//初始化dp数组dp[0][0] = grid[0][0];for(int i=1; i<col; i++){dp[0][i] = grid[0][i] + dp[0][i-1];}for(int i=1; i<row; i++){dp[i][0] = grid[i][0] + dp[i-1][0];}//遍历赋值dp数组for(int i=1; i<row; i++){for(int j=1; j<col; j++){dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1]) + grid[i][j];}}return dp[row-1][col-1];}
}