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.

code

import org.junit.Test;

public class MinimumPath {
    //矩阵的最小路径和:从矩阵的左上角到右下角的最小路径和，每次只能向左和向下移动
    public int minPathSum(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) return 0;
        int m = grid.length, n = grid[0].length;
        int[] dp = new int[n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (j == 0) dp[0] = dp[0] + grid[i][0];           // 只能从上侧走到该位置
                else if (i == 0) dp[j] = dp[j - 1] + grid[0][j];  // 只能从右侧走到该位置
                else dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
            }
        }
        return dp[n - 1];
    }
    @Test
    public void test(){
        int[][] grid={{1,3,1},{1,5,1},{4,2,1}};
        System.out.println(minPathSum(grid));
            }
}