由于在这个问题,涉及到求解依赖重叠的问题
即求到最后一个格子的最优路径取决于到最后一个格子的“正上方”、“正左方”两个格子的最优路径
因此果断使用动态规划
子问题为 到格子[I][J]的最优路径
状态转移方程请见代码注释
class Solution:
def __init__(self):
self.newgrid = []
self.grid = []
def minPathSum(self, grid: List[List[int]]) -> int:
m = len(grid)#行数
n = len(grid[0])#列数
if m == 1 and n == 1:
return grid[0][0]
elif (m == 1 and n != 1):
return sum(grid[0])
elif (m != 1 and n == 1):
res = 0
for i in range(m):
res += grid[i][0]
return res
#以下m和n均大于1
self.grid = grid
newlist = []
for i in range(m):
templist = []
for j in range(n):
templist.append(-1)
newlist.append(templist)
#以下为状态转移方程
for i in range(m):
#templist = []
for j in range(n):
if i == 0 and j == 0:#dp[0][0] = grid[0][0]
newlist[i][j] = grid[0][0]
elif i == 0 and j != 0:#对第一行,dp[0][j] = dp[0][j-1] + grid[0][j]
newlist[i][j] = newlist[0][j-1] + grid[0][j]
elif i != 0 and j == 0:#对第一列,dp[i][0] = dp[i-1][0] + grid[i][0]
newlist[i][j] = newlist[i-1][0] + grid[i][0]
for i in range(1,m):
for j in range(1,n):
#对其它位置的一般格子(不位于左边界和上边界上的格子)
#有dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j]
newlist[i][j] = min(newlist[i-1][j], newlist[i][j-1]) + grid[i][j]
return newlist[m-1][n-1]