diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index 23357f21..80c35c03 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -119,6 +119,19 @@ class Solution: return False ``` +```python +## for循环 +class Solution: + def canJump(self, nums: List[int]) -> bool: + cover = 0 + if len(nums) == 1: return True + for i in range(len(nums)): + if i <= cover: + cover = max(i + nums[i], cover) + if cover >= len(nums) - 1: return True + return False +``` + ### Go ```Go func canJUmp(nums []int) bool { diff --git a/problems/0121.买卖股票的最佳时机.md b/problems/0121.买卖股票的最佳时机.md index 63ac5d04..cf17f48d 100644 --- a/problems/0121.买卖股票的最佳时机.md +++ b/problems/0121.买卖股票的最佳时机.md @@ -310,6 +310,18 @@ class Solution: return dp[(length-1) % 2][1] ``` +> 动态规划:版本三 +```python +class Solution: + def maxProfit(self, prices: List[int]) -> int: + length = len(prices) + dp0, dp1 = -prices[0], 0 #注意这里只维护两个常量,因为dp0的更新不受dp1的影响 + for i in range(1, length): + dp1 = max(dp1, dp0 + prices[i]) + dp0 = max(dp0, -prices[i]) + return dp1 +``` + Go: > 贪心法: ```Go diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 214ff311..f9d3f97f 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -264,6 +264,25 @@ class Solution: return max(dp[-1][0], dp[-1][1]) ``` +```python +class Solution: + def wiggleMaxLength(self, nums: List[int]) -> int: + # up i作为波峰最长的序列长度 + # down i作为波谷最长的序列长度 + n = len(nums) + # 长度为0和1的直接返回长度 + if n<2: return n + for i in range(1,n): + if nums[i]>nums[i-1]: + # nums[i] 为波峰,1. 前面是波峰,up值不变,2. 前面是波谷,down值加1 + # 目前up值取两者的较大值(其实down+1即可,可以推理前一步down和up最多相差1,所以down+1>=up) + up = max(up, down+1) + elif nums[i]= minPrice and prices[i] <= minPrice + fee: - continue - else: - result += prices[i] - minPrice - fee + elif prices[i] > (minPrice + fee): # 此时有利润,同时假买入高价的股票,看看是否继续盈利 + result += prices[i] - (minPrice + fee) minPrice = prices[i] - fee + else: # minPrice<= prices[i] <= minPrice + fee, 价格处于minPrice和minPrice+fee之间,不做操作 + continue return result ```