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Commit 64845f4

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add new script
1 parent 0f90258 commit 64845f4

5 files changed

+324
-0
lines changed
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# 给定整数数组 nums 和整数 k,请返回数组中第 k 个最大的元素。
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#
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# 请注意,你需要找的是数组排序后的第 k 个最大的元素,而不是第 k 个不同的元素。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: [3,2,1,5,6,4], k = 2
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# 输出: 5
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#
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#
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# 示例 2:
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#
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#
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# 输入: [3,2,3,1,2,4,5,5,6], k = 4
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# 输出: 4
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= k <= nums.length <= 105
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# -104 <= nums[i] <= 104
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#
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# Related Topics 数组 分治 快速选择 排序 堆(优先队列)
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# 👍 1787 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def findKthLargest(self, nums: List[int], k: int) -> int:
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def partition(nums, left, right):
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key = left
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while left<right:
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while left<right and nums[key]<=nums[right]:
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right-=1
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while left<right and nums[left]<=nums[key]:
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left+=1
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nums[left], nums[right] = nums[right],nums[left]
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nums[left], nums[key] = nums[key], nums[left]
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return left
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def sort(nums, left, right):
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if left>=right:
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return
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mid = partition(nums, left, right)
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if mid==k:
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return
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elif mid<k:
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sort(nums, mid + 1, right)
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else:
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sort(nums, left, mid-1)
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l = len(nums)
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k = l-k
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sort(nums, 0, len(nums)-1)
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return nums[k]
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[28]实现 strStr().py

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# 实现 strStr() 函数。
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#
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# 给你两个字符串 haystack 和 needle ,请你在 haystack 字符串中找出 needle 字符串出现的第一个位置(下标从 0 开始)。如
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# 果不存在,则返回 -1 。
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#
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# 说明:
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#
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# 当 needle 是空字符串时,我们应当返回什么值呢?这是一个在面试中很好的问题。
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#
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# 对于本题而言,当 needle 是空字符串时我们应当返回 0 。这与 C 语言的 strstr() 以及 Java 的 indexOf() 定义相符。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:haystack = "hello", needle = "ll"
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# 输出:2
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#
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#
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# 示例 2:
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#
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#
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# 输入:haystack = "aaaaa", needle = "bba"
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# 输出:-1
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= haystack.length, needle.length <= 104
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# haystack 和 needle 仅由小写英文字符组成
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#
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# Related Topics 双指针 字符串 字符串匹配
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# 👍 1520 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def strStr(self, haystack: str, needle: str) -> int:
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def getnext(needle):
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l = len(needle)
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ptr1 = 0
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ptr2 = -1
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next = [-1] * l #相同前缀长度
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while ptr1<(l-1):
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if ptr2==-1 or needle[ptr1] ==needle[ptr2]:
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# 每一步更新的是下一个字符的前缀长度
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ptr1+=1 #下一个字符
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ptr2+=1 # 从哪一个位置开始匹配
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next[ptr1] = ptr2
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else:
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# 上一个字符匹配到的位置重新开始匹配
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ptr2 = next[ptr2]
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return next
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next = getnext(needle)
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l1 = len(haystack)
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l2 = len(needle)
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ptr1 =0
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ptr2 =0
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while ptr1<l1 and ptr2 <l2:
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if ptr2==-1 or haystack[ptr1] == needle[ptr2]:
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ptr1+=1
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ptr2+=1
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else:
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ptr2 = next[ptr2]
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if ptr2==l2:
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return ptr1-ptr2
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else:
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return -1
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# leetcode submit region end(Prohibit modification and deletion)
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# 给定一个整数数组 prices,其中第 prices[i] 表示第 i 天的股票价格 。
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#
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# 设计一个算法计算出最大利润。在满足以下约束条件下,你可以尽可能地完成更多的交易(多次买卖一支股票):
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#
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#
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# 卖出股票后,你无法在第二天买入股票 (即冷冻期为 1 天)。
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#
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#
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# 注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: prices = [1,2,3,0,2]
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# 输出: 3
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# 解释: 对应的交易状态为: [买入, 卖出, 冷冻期, 买入, 卖出]
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#
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# 示例 2:
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#
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#
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# 输入: prices = [1]
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# 输出: 0
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= prices.length <= 5000
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# 0 <= prices[i] <= 1000
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#
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# Related Topics 数组 动态规划
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# 👍 1272 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def maxProfit(self, prices: List[int]) -> int:
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#状态买入,T卖出,T-1卖出,T-2卖出
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l = len(prices)
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b = [0] * l
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s1 = [0] * l # 当天卖出
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s2 = [0] * l # T-1卖出
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s3 = [0] * l # T-2之前卖出
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b[0] = -prices[0]
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for i in range(1, l):
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b[i] = max(b[i-1], s3[i-1]-prices[i], s2[i-1]-prices[i])
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s1[i] = b[i-1]+prices[i]
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s2[i] = s1[i-1]
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s3[i] = max(s3[i-1], s2[i-1])
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return max(s1[-1],s2[-1],s3[-1])
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[347]前 K 个高频元素.py

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# 给你一个整数数组 nums 和一个整数 k ,请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: nums = [1,1,1,2,2,3], k = 2
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# 输出: [1,2]
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#
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#
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# 示例 2:
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#
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#
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# 输入: nums = [1], k = 1
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# 输出: [1]
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= nums.length <= 105
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# k 的取值范围是 [1, 数组中不相同的元素的个数]
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# 题目数据保证答案唯一,换句话说,数组中前 k 个高频元素的集合是唯一的
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#
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#
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#
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#
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# 进阶:你所设计算法的时间复杂度 必须 优于 O(n log n) ,其中 n 是数组大小。
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# Related Topics 数组 哈希表 分治 桶排序 计数 快速选择 排序 堆(优先队列)
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# 👍 1275 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def topKFrequent(self, nums: List[int], k: int) -> List[int]:
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import collections
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counter = collections.Counter(nums)
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counter = list(counter.items())
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def partition(nums, left, right):
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key= left
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while left<right:
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while left<right and nums[key][1]<=nums[right][1]:
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right-=1
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while left<right and nums[key][1]>=nums[left][1]:
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left+=1
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nums[left], nums[right] = nums[right], nums[left]
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nums[left], nums[key] =nums[key], nums[left]
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return left
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def sort(nums, k, left, right):
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if left>=right:
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return
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mid = partition(nums, left, right)
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if mid == k:
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return
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elif mid >k:
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sort(nums, k, left, mid - 1)
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else:
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sort(nums, k, mid+1, right)
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l = len(counter)
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sort(counter, l-k, 0, l-1)
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return [i[0] for i in counter[-k:]]
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[605]种花问题.py

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# 假设有一个很长的花坛,一部分地块种植了花,另一部分却没有。可是,花不能种植在相邻的地块上,它们会争夺水源,两者都会死去。
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#
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# 给你一个整数数组 flowerbed 表示花坛,由若干 0 和 1 组成,其中 0 表示没种植花,1 表示种植了花。另有一个数 n ,能否在不打破种植规则
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# 的情况下种入 n 朵花?能则返回 true ,不能则返回 false。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:flowerbed = [1,0,0,0,1], n = 1
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# 输出:true
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#
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#
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# 示例 2:
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#
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#
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# 输入:flowerbed = [1,0,0,0,1], n = 2
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# 输出:false
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= flowerbed.length <= 2 * 104
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# flowerbed[i] 为 0 或 1
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# flowerbed 中不存在相邻的两朵花
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# 0 <= n <= flowerbed.length
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#
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# Related Topics 贪心 数组
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# 👍 467 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
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prepos = -1
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counter = 0
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l = len(flowerbed)
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for i in range(l):
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if flowerbed[i]==1:
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if prepos<0:
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counter += (i-prepos-1)//2
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else:
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counter += (i-prepos-2)//2
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prepos = i
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if prepos<0:
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counter+=(l-prepos)//2
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else:
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counter+=(l-prepos-1)//2
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print(counter)
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if counter>=n:
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return True
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else:
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return False
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# leetcode submit region end(Prohibit modification and deletion)

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