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Commit 204f091

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🐱(offer): 面试题 17.16. 按摩师
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‎docs/data-structure/array/README.md‎

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@@ -2630,6 +2630,49 @@ func getMin(a int, b int) int {
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}
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```
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## 945. 使数组唯一的最小增量
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[原题链接](https://leetcode-cn.com/problems/minimum-increment-to-make-array-unique/)
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### 解一:计数
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一开始暴力破解超时了,优化一下采用以下写法:
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1. 先把重复出现的数字存储起来
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2. 遍历 0~80000 的数字(最坏情况出现 40000 个 40000,最多可以叠加到 80000)
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- 如果发现数字是重复出现过的:
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- 将记录重复出现数字的 `repeat` 变量 + 1
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- 在结果 `res` 提前减去 `重复个数 * 重复数字`
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- 如果发现是空的位置:
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- 用空的位置处理一个重复出现的数字:`repeat -= 1`
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- 在 `res` 加上空位的数字
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```python
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class Solution:
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def minIncrementForUnique(self, A: List[int]) -> int:
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count = [0 for _ in range(80000)]
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for x in A:
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count[x] += 1
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# 总数
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res = 0
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# 重复数量
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repeat = 0
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for i in range(80000):
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if count[i] > 1:
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# 出现重复
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repeat += count[i] - 1
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# 减去多余的数
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res -= i * (count[i] - 1)
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# 没有出现过的数
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if count[i] == 0 and repeat > 0:
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repeat -= 1 # 处理一个重复的数
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res += i
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return res
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```
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<!-- tabs:end -->
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## 1002. 查找常用字符

‎docs/offer/README.md‎

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- 时间复杂度:$O(n)$
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- 空间复杂度:$O(n)$
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## 面试题 17.16. 按摩师
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[原题链接](https://leetcode-cn.com/problems/the-masseuse-lcci/)
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### 动态规划
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- 用 `dp[0][i]` 表示不接 i 预约可以获得的最长预约时间
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- 用 `dp[1][i]` 表示接 i 预约可以获得的最长预约时间
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```python
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class Solution:
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def massage(self, nums: List[int]) -> int:
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# dp[0][i] 不接
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# dp[1][i] 接
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length = len(nums)
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if length == 0:
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return 0
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dp = [[0 for _ in range(length)] for _ in range(2)]
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# 转移方程:dp[0][i] = max(dp[1][i - 1], dp[0][i - 1])
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# dp[1][i] = max(dp[0][i - 1] + nums[i])
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dp[1][0] = nums[0]
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for i in range(1, length):
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dp[0][i] = max(dp[1][i - 1], dp[0][i - 1])
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dp[1][i] = dp[0][i - 1] + nums[i]
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return max(dp[0][length - 1], dp[1][length - 1])
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```
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## 面试题24. 反转链表
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[原题链接](https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/)

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