122.买卖股票的最佳时机 II
1、贪心
class Solution:
def maxProfit(self, prices: List[int]) -> int:
res = 0
for i in range(1, len(prices)):
res += max(prices[i]-prices[i-1], 0)
return res
2、动态规划
class Solution:
def maxProfit(self, prices: List[int]) -> int:
n = len(prices)
# 第一列代表第几天，第二列代表是否持有股票
dp = [[0] * 2 for _ in range(n)]
# 0 代表持有股票， 1 代表不持有股票
dp[0][0] = -prices[0]
dp[0][1] = 0
for i in range(1, n):
dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i])
dp[i][1] = max(dp[i-1][1], dp[i-1][0]+prices[i])
return dp[-1][1]
class Solution:
def canJump(self, nums: List[int]) -> bool:
cover = 0 # 每个位置可以覆盖的范围
n = len(nums)
if n == 1:
return True
for i in range(n):
if i <= cover:
cover = max(cover, nums[i]+i)
if cover >= n-1:
return True
return False
45.跳跃游戏 II
1、走一步看一步
class Solution:
def jump(self, nums: List[int]) -> int:
n = len(nums)
if n == 1:
return 0
cur_distance = 0 # 当前覆盖的最远距离
res = 0 # 移动的步数
next_distance = 0 # 下一步覆盖的最远距离
for i in range(n):
next_distance = max(nums[i]+i, next_distance)
if i == cur_distance:
res += 1 # 需要移动下一步
cur_distance = next_distance # 更新当前可以移动的最远距离
if next_distance >= n - 1:
break
return res
2、动态规划
class Solution:
def jump(self, nums: List[int]) -> int:
n = len(nums)
# 初始化数组, 每个位置需要多少步到
dp = [10**4+1] * n
dp[0] = 0
for i in range(n): # 遍历数组
for j in range(nums[i]+1): # 遍历覆盖范围
if i + j < n: # 避免走出数组范围
dp[i+j] = min(dp[i+j], dp[i]+1)
return dp[-1]