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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.
You can only move either down or right at any point in time!
Dynamic programming is ultilized to solve this problem.
First of all, define another matrix which has same dimension for both x and y. And let us define the number stored in the array stand for the minimum summation of all the path to the position with same x and y in grid matrix.
Then the sum[i][j] = min(sum[i-1][j],sum[i][j-1]) + grid[i][j];
Initialize the sum[0][0] = grid [0][0]; initialize the two boundaries with all the summation of previous path to that certain node.
Solve sum[m-1][n-1]
1 public class Solution { 2 /** 3 * @param grid: a list of lists of integers. 4 * @return: An integer, minimizes the sum of all numbers along its path 5 */ 6 public int minPathSum(int[][] grid) { 7 // write your code here 8 if (grid == null || grid.length ==0 || grid[0].length == 0) { 9 return 0; 10 } 11 int m = grid.length; 12 int n = grid[0].length; 13 for(int i = 1; i < m; i++) { 14 grid[i][0] = grid[i-1][0] + grid[i][0]; 15 } 16 for(int i = 1; i < n; i++) { 17 grid[0][i] = grid[0][i-1] + grid[0][i]; 18 } 19 for(int i= 1; i < m; i ++) { 20 for (int j =1; j < n; j++) { 21 grid[i][j] = Math.min(grid[i-1][j],grid[i][j-1]) + grid[i][j]; 22 } 23 } 24 return grid[m-1][n-1]; 25 } 26 }
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原文地址：http://www.cnblogs.com/ly91417/p/5760450.html
