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.
[分析] 倒推思路,从后往前推,要求走到最后一个格子的minPath, 若知道所有可能的倒数第二个格子的minPath,取其小者加上最后一格的值即可,题中之前一个只能上相邻的上面一格和左边一格,如此可得递推公式。
public class Solution { // dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; public int minPathSum(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) return 0; int m = grid.length; int n = grid[0].length; int[][] dp = new int[m][n]; dp[0][0] = grid[0][0]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (i > 0 && j > 0) dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; else if (i > 0) dp[i][j] = dp[i - 1][j] + grid[i][j]; else if (j > 0) dp[i][j] = dp[i][j - 1] + grid[i][j]; } } return dp[m - 1][n - 1]; } }