代码随想录算法训练营第二十八天|LeetCode 122 买卖股票的最佳时机 II，LeetCode 55 跳跃游戏，LeetCode 45 跳跃游戏 II，LeetCode 1005-优快云博客

1.LeetCode 122 买卖股票的最佳时机 II

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        result = 0
        for i in range(1, len(prices)):
            result += max(prices[i] - prices[i - 1], 0)
        return result

第一题结束

2.LeetCode 55 跳跃游戏

题目链接：55. 跳跃游戏

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

第二题结束

3.LeetCode 45 跳跃游戏 II

题目链接：45. 跳跃游戏 II

class Solution:
    def jump(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        
        count = 0
        i = 0
        cover = 0

        while i <= cover:
            for i in range(i, cover + 1):
                cover = max(i + nums[i], cover)
                if cover >= len(nums) - 1:
                    return count + 1
            count += 1

第三题结束

4.LeetCode 1005 K次取反后最大化的数组和

题目链接：1005. K 次取反后最大化的数组和

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        nums.sort(key=lambda x: abs(x), reverse=True)
        for i in range(len(nums)):
            if nums[i] < 0 and k > 0:
                nums[i] *= -1
                k -= 1
        if k % 2 == 1:
            num[-1] *= -1
        return sum(nums)

第四题结束

今天用时1.5h，贪心有点难