LeetCode刷题日记 - 持续更新

740. 删除并获得点数

Note: 进阶版本打家劫舍，考虑nums[i], nums[i - 1], nums[i + 1]，即相邻不能“偷”，原始数组通过累加得到新数组；对新数组进行动态规划即可；
代码可能有点小瑕疵，但是能够AC

class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        max_num = max(nums)
        num_list = [0] * (max_num + 1)
        for num in nums:
            num_list[num] += num

        dp = [0] * len(num_list)
        for i in range(max_num + 1):
            if i == 0:
                dp[i] = num_list[i]
            else:
                dp[i] = max(dp[i - 1], dp[i - 2] + num_list[i])
        return dp[max_num]

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746. 使用最小花费爬楼梯

Note: 因为最小花费，则动态规划方程中，前一次可以选择不爬，前前一次则必须爬到当前位置，所以方程为： dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        dp = [0] * n
        for i in range(n):
            if i < 2:
                dp[i] = cost[i]
            else:
                dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
        return min(dp[n - 1], dp[n - 2])

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62. 不同路径

Note: 初始i == 0 or j == 0表示在边界，仅有一种方法能到达，因此初始化为1；

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if i == 0:
                    dp[i][j] = 1
                elif j == 0:
                    print(i, j)
                    dp[i][j] = 1
                else:
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

        return dp[m - 1][n - 1]