[leetcode64]Minimum Path Sum

问题描述：

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.

算法：

我首先想到的是利用优先队列进行搜索。

import heapq

class Solution(object):

def minPathSum(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
endx = len(grid) - 1
endy = len(grid[0]) - 1
priority_que = [(grid[0][0], (0,0))]
while True:
curSum, point = heapq.heappop(priority_que)
if point == (endx,endy):
return curSum
curX,curY = point
successors = [(x,y)for (x,y) in [(curX+1, curY), (curX, curY+1)] if x<=endx and y<=endy]
for x,y in successors:
if (x,y) == (endx,endy):
return curSum+grid[x][y]
else:
heapq.heappush(priority_que, (curSum+grid[x][y],(x,y)))

但是，提交代码后发现超时。

这时，首先想到的是动态规划。

状态转移方程为 func(x,y) = min(func(x-1,y),func(x,y-1))+grid(x,y)

使用矩阵存储从起点到当前位置的路径最小值，从最上面一层以及最左面一层，逐渐递推到结束位置。

class Solution(object):

def minPathSum(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
endx = len(grid)
endy = len(grid[0])
path = [[0 for _ in range(endy)] for _ in range(endx)]
path[0][0] = grid[0][0]
# get up line ready
for i in range(1, endy):
path[0][i] = path[0][i-1] + grid[0][i]
# get left line ready
for i in range(1, endx):
print(path)
print(i)
path[i][0] = path[i-1][0] + grid[i][0]

for r in range(1, endx):
for c in range(1, endy):
path[r][c] = min(path[r-1][c], path[r][c-1]) + grid[r][c]
return path[endx-1][endy-1]