您的位置:首页 > 编程语言 > Python开发

[置顶] LeetCode--Minimum Path Sum (最小路径和)Python

2017-11-17 11:19 375 查看
题目:

给定一个m*n的数组,数组中包含非负数,从该数组左上角到该数组右下角的最小路径和。(只能向下或者向右移动)。数组格式如下:

[[1,3,1],

[1,5,1],

[4,2,1]]

解题思路:

考虑使用递归或者动态规划来解题:

楼主刚开始用递归实现了一遍,发现超时了,改为动态规划后AC。故只解释动归代码:从左上角开始对数组进行遍历,将grid(数组)内容存储为走到当前位置的最短路径和。故只考虑当前位置的左边和上边哪个小,就选择哪个路径即可。

动态规划代码(Python):

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


递归代码(Python):

class Solution(object):
def minPathSum(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
final = []
m = len(grid)-1
n = len(grid[0])-1
print m
print n
def Iteration(i,j,sum_0):
if i<m and j<n:
Iteration(i+1,j,sum_0+grid[i+1][j])
Iteration(i,j+1,sum_0+grid[i][j+1])
if i==m and j==n:
final.append(sum_0)
if i==m and j<n:
Iteration(i,j+1,sum_0+grid[i][j+1])
if i<m and j==n:
Iteration(i+1,j,sum_0+grid[i+1][j])
Iteration(0,0,grid[0][0])
return min(final)
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: