leetcode 之 Minimum Path Sum

题目描述: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.给定一个m*n的矩形a, 计算从a[0][0]到a[m]
的每个数字加起来最小的和. 注意, 每次只能往下或者右查找下一个数字.分析:每次都往右下角寻找. 之前看到一道题, 计算n*n的方形从左上角到又下角最短的距离有多少种走法. 我们先来看这个问题.假设在4*4的格子中, a[i][j] 的值 等于a[i-1][j] + a[j-1][i], 如下图所示:1 1 1 11 2 3 41 3 6 101 4 10 20上面展示的是的最短路径有多少种走法.然后在计算从左上角到右下角最小数字和时,采用动态规划的思想, 每一步的最小取值,依赖其左边和上边的最小值,我们设m[i][j] 是走到第a[i][j]时,最小的元素和, 由上面可以得出,m[i][j] = min(m[i-1][j]+a[i][j], m[i][j-1]+a[i][j])其中: m[i][0] = a[i][0] m[0][j] = a[0][j]代码如下:

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