64. Minimum Path Sum
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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]
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