62.不同路径
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
# dp[i][j] 代表到达 dp[i][j] 有多少不同路径
dp = [[0]*n for _ in range(m)]
# 初始化
for i in range(m):
dp[i][0] = 1
for j in range(n):
dp[0][j] = 1
# 遍历
for i in range(1, m):
for j in range(1, n):
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[m-1][n-1]
63. 不同路径 II
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m, n = len(obstacleGrid), len(obstacleGrid[0])
if obstacleGrid[0][0] == 1 or obstacleGrid[m-1][n-1] == 1:
return 0
dp = [[0]*n for _ in range(m)]
for i in range(m):
if obstacleGrid[i][0] == 0:
dp[i][0] = 1
else:
break
for j in range(n):
if obstacleGrid[0][j] == 0:
dp[0][j] = 1
else:
break
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j] == 1:
continue
else:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[m-1][n-1]
343. 整数拆分
class Solution:
def integerBreak(self, n: int) -> int:
# dp[i] 代表拆分整数 i 获取的最大值
dp = [0] * (n+1)
# 初始化 dp 数组
dp[2] = 1
# 遍历 dp 数组
for i in range(2, n+1):
# 从 1 开始拆分整数
for j in range(1, i//2 + 1):
dp[i] = max(dp[i], j*(i-j), j*dp[i-j])
return dp[n]