推荐常州模板网站建设关键词优化步骤简短
目录
1. 循环随机取数组直到得出指定数字？
2. 旋转链表
3. 区间和的个数
1. 循环随机取数组直到得出指定数字？
举个例子： 随机数字范围：0~100 每组数字量：6（s1,s2,s3,s4,s5,s6） 第二轮开始随机数字范围：新s1和新s2取值为旧s1和s2之间，新s3和新s4取值为旧s3和s4之间，新s5和新s6取值为旧s5和s6之间。 跳出循环条件：任意数字=37 如因s1=s2!=37&&s3=s4!=37&&s5=s6!=37使数组进入无意义无限循环，则重新取0~100六个数字并开始如上述第二轮随机的随机取值。
代码：
import randomdef random_test():rst_list = [random.randint(0,100) for i in range(0, 6)]print(rst_list)while 1:temp = []for k,v in enumerate(rst_list):if k%2==0:temp.append(random.randint(min([rst_list[k],rst_list[k+1]]),max([rst_list[k],rst_list[k+1]])))else:temp.append(random.randint(min(rst_list[k-1], rst_list[k]),max(rst_list[k-1], rst_list[k])))rst_list = tempprint(rst_list)if 37 in rst_list:print("rst_list:",rst_list)return rst_listelse:if rst_list[0]==rst_list[1] and rst_list[2]==rst_list[3] and rst_list[4]==rst_list[5]:rst_list = [random.randint(0, 100) for i in range(0, 6)]def main():random_test()if __name__ == "__main__":main()[20, 2, 33, 26, 67, 7]
[15, 6, 28, 28, 8, 29]
[15, 13, 28, 28, 29, 21]
[13, 14, 28, 28, 22, 27]
[14, 13, 28, 28, 27, 24]
[14, 13, 28, 28, 27, 25]
[14, 14, 28, 28, 27, 26]
[14, 14, 28, 28, 27, 27]
[51, 38, 46, 43, 45, 41]
[42, 39, 43, 43, 43, 43]
[39, 40, 43, 43, 43, 43]
[40, 39, 43, 43, 43, 43]
[40, 39, 43, 43, 43, 43]
[39, 39, 43, 43, 43, 43]
[91, 22, 53, 47, 20, 20]
[40, 76, 53, 47, 20, 20]
[44, 68, 48, 49, 20, 20]
[65, 51, 49, 48, 20, 20]
[58, 57, 49, 49, 20, 20]
[57, 57, 49, 49, 20, 20]
[37, 42, 15, 19, 61, 63]
rst_list: [37, 42, 15, 19, 61, 63]#注： 随机数字输出
2. 旋转链表
给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。
示例 1：
输入：head = [1,2,3,4,5], k = 2 输出：[4,5,1,2,3]
示例 2：
输入：head = [0,1,2], k = 4 输出：[2,0,1]
提示：
- 链表中节点的数目在范围
[0, 500]内 -100 <= Node.val <= 1000 <= k <= 2 * 109
代码：
class ListNode(object):def __init__(self, x):self.val = xself.next = Noneclass LinkList:def __init__(self):self.head=Nonedef initList(self, data):self.head = ListNode(data[0])r=self.headp = self.headfor i in data[1:]:node = ListNode(i)p.next = nodep = p.nextreturn rdef convert_list(self,head):ret = []if head == None:returnnode = headwhile node != None:ret.append(node.val)node = node.nextreturn retclass Solution(object):def rotateRight(self, head, k):""":type head: ListNode:type k: int:rtype: ListNode"""if not head or k == 0:return headslow = fast = headlength = 1while k and fast.next:fast = fast.nextlength += 1k -= 1if k != 0:k = (k + length - 1) % length return self.rotateRight(head, k)else:while fast.next:fast = fast.nextslow = slow.nextreturn self.rotate(head, fast, slow)def rotate(self, head, fast, slow):fast.next = headhead = slow.nextslow.next = Nonereturn head# %%
l = LinkList()
list1 = [0,1,2]
k = 4
l1 = l.initList(list1)
s = Solution()
print(l.convert_list(s.rotateRight(l1, k)))
输出：
[2, 0, 1]
3. 区间和的个数
给你一个整数数组 nums 以及两个整数 lower 和 upper 。求数组中，值位于范围 [lower, upper] （包含 lower 和 upper）之内的 区间和的个数 。
区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。
示例 1：
输入：nums = [-2,5,-1], lower = -2, upper = 2 输出：3 解释：存在三个区间：[0,0]、[2,2] 和 [0,2] ，对应的区间和分别是：-2 、-1 、2 。
示例 2：
输入：nums = [0], lower = 0, upper = 0 输出：1
提示：
1 <= nums.length <= 105-231 <= nums[i] <= 231 - 1-105 <= lower <= upper <= 105- 题目数据保证答案是一个 32 位 的整数
代码：
class Solution:def countRangeSum(self, nums: list, lower: int, upper: int) -> int:prefix = [0]last = 0for num in nums:last += numprefix.append(last)def count(pp, left, right):if left == right:return 0else:mid = (left + right) // 2ans = count(pp, left, mid) + count(pp, mid + 1, right)i1, i2, i3 = left, mid + 1, mid + 1while i1 <= mid:while i2 <= right and pp[i2] - pp[i1] < lower:i2 += 1while i3 <= right and pp[i3] - pp[i1] <= upper:i3 += 1ans += i3 - i2i1 += 1result = []i1, i2 = left, mid + 1while i1 <= mid and i2 <= right:if pp[i1] <= pp[i2]:result.append(pp[i1])i1 += 1else:result.append(pp[i2])i2 += 1while i1 <= mid:result.append(pp[i1])i1 += 1while i2 <= right:result.append(pp[i2])i2 += 1for i in range(len(result)):pp[i + left] = result[i]return ansreturn count(prefix, 0, len(prefix) - 1)if __name__ == "__main__":s = Solution()print(s.countRangeSum([-2,5,-1],-2,2))print(s.countRangeSum([0],0,0))输出：
3
1