64. Minimum Path Sum

64. Minimum Path Sum

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

2. 思路

本题的要求是从矩阵的左上角到右下角找出总和最小的路径。那么什么样的路径总和最小呢。如果走到了第i行第j列要是总和最小，则要求能够到达第i行第j列的前一步的总和最小：

dp[i] [j] = grid[i] [j] + min(dp[i-1] [j],dp[i] [j-1])

3. 实现

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        row = len(grid)
        col = len(grid[0])
        dp = [ [ 0 for j in range(0,col)] for i in range(0,row)  ]
        
        dp[0][0] = grid[0][0]
        for i in range(1,row):
            dp[i][0] = dp[i-1][0] + grid[i][0]
            
        for j in range(1,col):
            dp[0][j] = dp[0][j-1] + grid[0][j]
            
        for i in range( 1, row ):
            for j in range( 1, col):
                dp[i][j] = grid[i][j] + min(dp[i-1][j],dp[i][j-1])
        
        return dp[row-1][col-1]