No.154 FindMin 寻找旋转排序数组中的最小值 II 题目 假设按照升序排序的数组在预先未知的某个点上进行了旋转. ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] ). 请找出其中最小的元素. 注意数组中可能存在重复的元素. 示例 输入: [1,3,5] 输出: 1 输入: [2,2,2,0,1] 输出: 0 说明 这道题是 No.153 寻找旋转排序数组中的最小值 的延伸题目. 允许重复会影响算法的时间复杂度吗?会如何影响,为什么? 思路 代…

[LeetCode]154. Find Minimum in Rotated Sorted Array II 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/description/ 题目描述: Follow up for "Find Minimum in Rotated Sorted Array": What if duplicat…

下面题目是LeetCode算法155题: https://leetcode.com/problems/min-stack/ 题目1:最小函数min()栈 设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈. push(x) -- 将元素 x 推入栈中. pop() -- 删除栈顶的元素. top() -- 获取栈顶元素. getMin() -- 检索栈中的最小元素. package com.good.good.study.stack; import java.uti…

题目 Follow up for "Find Minimum in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 migh…

#-*- coding: UTF-8 -*- class MinStack(object):    def __init__(self):        """        initialize your data structure here.        """        self.Stack=[]        self.minStack=[]            def push(self, x):        "&…

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be…

Find Minimum in Rotated Sorted Array II Follow up for "Find Minimum in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you…

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. getMin() -- Retrieve the minimum e…

估计要刷很久才能突破三道题了.还是刷的太少.尽管对了前两题,但是我觉得写的不怎么样.还是将所有题目都写一下吧. 5189. "气球" 的最大数量 题目比较简单.就是找出一个字符串中,balloon 中每个字母出现的次数,次数最小的就是结果.注意,l 和 o 要除以 2. 定义一个数组统计每个字母的个数. int[] count = new int[26];// 统计每个字母的个数 时间复杂度:只有一个循环,故为 O(n)O(n)O(n). 空间复杂度:只有一个一维数组大小恒为 26,故…

一."气球" 的最大数量(LeetCode-5189) 1.1 题目描述 1.2 解题思路 统计各个字母的出现的次数,然后根据"木桶最短板"返回就好. 1.3 解题代码 public class Solution { public int maxNumberOfBalloons(String text) { int[] arr = new int[5]; char[] charArray = text.toCharArray(); for(char ch:charA…