64. Minimum Path Sum

QUESTION

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.

THOUGHT I

对矩阵的所有路径进行遍历，保存最小的路径和。

CODE

public class Solution {    int minSum = Integer.MAX_VALUE;    public int minPathSum(int[][] grid) {        if(grid.length == 0 || grid[0].length == 0)            return 0;        return searchMinSum(grid,0,0,0);    }    public int searchMinSum(int[][] grid,int x,int y,int sum){        if(x == grid.length - 1 && y == grid[0].length - 1){            sum += grid[grid.length - 1][grid[0].length - 1];            if(sum < minSum)                minSum = sum;        }        else{            if(y < grid[0].length - 1)                searchMinSum(grid,x,y + 1,sum + grid[x][y]);            if(x < grid.length - 1)                searchMinSum(grid,x + 1,y,sum + grid[x][y]);        }        return minSum;    }}

RESULT

超时了。

THOUGHT II

看到一种很巧妙的办法，貌似矩阵的问题好多都是这么解决的。首先把矩阵中的点分为四类类，起点，第一行中的点，第一列中的点，普通点；对于起点来说，保持不变。行中的每一点都为同一行前一列中的值和本点值的和。第一列中的点中的值变为同一列，前一行和本行的和。算法原理也比较好理解，因为路径只能往下走或者往右走，那么作为普通点，它的和来自于左边的点或者是上面的点，那么只要取最小的就可以了，这是动态规划的思想。而作为第一行或者第一列中的点，它的和只能是来自于左边或者上面的点。

CODE

public class Solution {    public int minPathSum(int[][] grid) {        int row = grid.length,col = grid[0].length;        for(int i = 0;i < row;i++){            for(int j = 0;j < col;j++){                if(i ==0 && j == 0)                    continue;                else if(i == 0)                    grid[i][j] += grid[i][j - 1];                else if(j == 0)                    grid[i][j] += grid[i -1][j];                else                    grid[i][j] += Math.min(grid[i - 1][j],grid[i][j - 1]);            }        }        return grid[row - 1][col - 1];    }}

RESULT

runtime complexity is O(n^2),space complexity is O(1);