Note: You can only move either down or right at any point in time.
最小路径求和的问题,用动态规划来解决。每经过一个点都记录从开始到这个点的最小带权路径。当i = 0 或者j = 0的时候,因为只有一个方向可以走,这时的递推式为grid[i][0] = grid[i - 1][0] + grid[i][0] (j = 0) 和grid[0][j] = grid[0][j - 1] + grid[0][j] ( i = 0)。当i >0 并且j > 0 时,递推式为grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j],每次都取一个和最小的路径。代码如下:
public class Solution { public int minPathSum(int[][] grid) { if(grid == null || grid.length == 0) return 0; int m = grid.length; int n = grid[0].length; for(int i = 1; i < m; i++) grid[i][0] = grid[i - 1][0] + grid[i][0]; for(int j = 1; j < n; j++) grid[0][j] = grid[0][j - 1] + grid[0][j]; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) { grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]; } return grid[m - 1][n - 1]; } }