标签：style blog io color ar sp for strong div
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.
Analysis:
This is a DP problem. d[i][j] is the min path sum from src to grid[i][j].
d[i][j] = min{d[i-1][j],d[i][j-1]}+grid[i][j];
Solution:
1 public class Solution { 2 public int minPathSum(int[][] grid) { 3 int xLen = grid.length; 4 if (xLen==0) return 0; 5 int yLen = grid[0].length; 6 if (yLen==0) return 0; 7 8 int[][] path = new int[xLen][yLen]; 9 path[0][0] = grid[0][0]; 10 for (int i = 1;i<yLen;i++) 11 path[0][i] = path[0][i-1]+grid[0][i]; 12 for (int i=1;i<xLen;i++) 13 path[i][0] = path[i-1][0]+grid[i][0]; 14 15 for (int i=1;i<xLen;i++) 16 for (int j=1;j<yLen;j++) 17 if (path[i-1][j]<path[i][j-1]) 18 path[i][j] = path[i-1][j]+grid[i][j]; 19 else path[i][j] = path[i][j-1]+grid[i][j]; 20 21 return path[xLen-1][yLen-1]; 22 23 } 24 }
NOTE: We can reduce the memory space to O(n) by using just on array. The formula:
d[i] = min{d[i],d[i-1]}+grid[i][j];
标签：style blog io color ar sp for strong div
原文地址：http://www.cnblogs.com/lishiblog/p/4102738.html
