64. Minimum Path Sum

题目

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.

思路

递归求解

代码

public class Solution {        //min = grid[i,j]+Min(minPath(grid,i-1,j),minPath(grid,i,j-1))        public int MinPathSum(int[,] grid)        {            int m = grid.GetUpperBound(0);            int n = grid.GetUpperBound(1);            return minPath(grid,m,n);        }        private int minPath(int[,] grid, int i, int j)        {            int min = grid[i, j];            if (i == 0 && j == 0) return grid[0, 0];            if (i == 0)                return min + Math.Min(minPath(grid, 0, j-1), minPath(grid, 0, j - 1));            if (j == 0)                return min + Math.Min(minPath(grid, i-1, 0), minPath(grid, i-1, 0));            return min+Math.Min(minPath(grid, i - 1, j), minPath(grid, i, j - 1));        }}

算法优化

不采用递归。

采用动态规划的思想，如下图所示，为了求得最短路径，要么是终点为h1的所有路径，或终点为h2的所有路径中的最小值，加上finish方框的值。

代码

        public int MinPathSum(int[,] grid)        {            int m=grid.GetUpperBound(0)+1; //0-dimension element size            int n=grid.GetUpperBound(1)+1; //1-dimension element size            int[,] sumdp = new int[m,n];            sumdp[0,0]=grid[0,0];            //two boundaries:            for (int i = 1; i < m; i++)                sumdp[i,0] = sumdp[i-1,0]+grid[i,0];            for (int i = 1; i < n; i++)                sumdp[0,i] = sumdp[0,i-1]+grid[0,i];            //dp[i][j] = dp[i-1][j]+dp[i][j-1]            for (int i = 1; i < m; i++)            {                for (int j = 1; j < n; j++)                {                    sumdp[i, j] = Math.Min(sumdp[i - 1, j], sumdp[i, j - 1]) + grid[i,j];                }            }            return sumdp[m - 1, n - 1];        }