LeetCode刷题记录2020-10-09之日常划水！！！

# 破坏链表数据、快慢指针都行
# 这里就放置一个大佬的优秀代码，学到了学到了
f = open("user.out", "w")
while True:
    try:
        param_1 = input()
        param_2 = int(input())
        f.write(‘true\n‘ if param_2>-1 else ‘false\n‘)
    except:
        break
exit()

def detectCycle(self, head: ListNode) -> ListNode:
        # 牺牲空间换时间
        myDict = {}
        index = 0
        while head:
            if myDict.get(head) is not None:
                return head
            else:
                myDict[head] = index
                index += 1
                head = head.next
        return None

https://leetcode-cn.com/problems/find-the-duplicate-number/
# return sum(nums) - int(len(nums) * (len(nums) - 1) / 2)
        # 桶记录
        visit = [-1 for _ in range(len(nums))]
        for n in nums:
            visit[n] += 1
            if visit[n] == 1:
                return n

def firstMissingPositive(self, nums: List[int]) -> int:
        # 哭了，做了半天题意没理解
        # min_n = float(‘inf‘)
        # max_n = -float(‘inf‘)
        # for n in nums:
        #     if n > 0:
        #         min_n = min(min_n, n)
        #     max_n = max(max_n, n)
        # visit = [0 for _ in range(max_n - min_n + 1)]
        # for n in nums:
        #     if n > 0:
        #         visit[n-min_n] = 1
        # for i in range(len(visit)):
        #     if visit[i] == 0:
        #         return min_n+i
        # return min_n - 1 if min_n > 1 else max_n + 1

        # O(4n) -> O(n)
        lenth = len(nums)
        if len(nums) == 0:
            return 1
        visit = [0 for _ in range(lenth+1)]
        for n in nums:
            if 1 <= n <= lenth:
                visit[n] = 1
        for i in range(1, len(visit)):
            if visit[i] == 0:
                return i
        return lenth+1