【代码随想录训练营】【Day 35】【贪心-2】| Leetcode 122, 55, 45
需强化知识点
- 贪心:跳跃游戏
题目
122. 买卖股票的最佳时机 II
- 动态规划
- 贪心:只要股票第二天涨了,前一天就买,第二就买
class Solution:def maxProfit(self, prices: List[int]) -> int:# dp = [[0] * 2 for _ in range(len(prices))]# # 0 代表当天持有,1 代表当天不持有# dp[0][0] = -prices[0]# result = 0# for i in range(1, len(prices)):# dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i])# dp[i][1] = max(dp[i-1][0]+prices[i], dp[i-1][1])# result = max(result, max(dp[i][0], dp[i][1]))# return resultresult = 0for i in range(1, len(prices)):if prices[i] > prices[i-1]:result += prices[i] - prices[i-1]return result
55. 跳跃游戏
- cover记录每步能覆盖到的最大范围,只要最终cover能到达最后一个下标,即为True,注意通过 if i <= cover:来控制,在cover范围内都可以更新cover
class Solution:def canJump(self, nums: List[int]) -> bool:cover = 0for i in range(len(nums)):if i <= cover:cover = max(cover, i+nums[i])if cover >= len(nums)-1:return Truereturn False
45. 跳跃游戏 II
- 最小跳跃数,即每一步都是最远
- 当到达目前cover,就要跳一步,选择跳到下一步能跳到的最远,因此new_cover要一直更新,维护着下一步能跳到的最远
class Solution:def jump(self, nums: List[int]) -> int:new_cover, cover = 0, 0result = 0if len(nums) == 1:return 0for i in range(len(nums)):new_cover = max(new_cover, nums[i] + i)if i == cover:cover = new_coverresult += 1if cover >= len(nums) - 1:return resultreturn result