leetcode Ch1-Search】的更多相关文章

索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 035. Search for a Range (Medium) 链接: 题目:https://leetcode.com/problems/search-for-a-range/ 代码(github):https://github.com/illuz/leetcode 题意: 在有序数组中找到一个数的范围.(由于数…
指数:[LeetCode] Leetcode 解决问题的指数 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 033. Search in Rotated Sorted Array (Hard) 链接: 题目:https://leetcode.com/problems/search-in-rotated-sorted-array/ 代码(github):https://github.com/illuz/leetcod…
leetcode - 35. Search Insert Position - Easy descrition Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the arr…
leetcode - 34. Search for a Range - Medium descrition Given an array of integers sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the ta…
leetcode - 33. Search in Rotated Sorted Array - Medium descrition Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search.…
LeetCode 81 Search in Rotated Sorted Array II [binary search] <c++> 给出排序好的一维有重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数是否存在,时间复杂度限制\(O(log_2n)\) C++ 因为有重复元素存在,nums[l] <= nums[mid]不能说明[l,mid]区间内一定是单调的,比如数组[1,2,3,1,1,1,1],但是严格小于和严格大于的情况还是可以…
LeetCode 33 Search in Rotated Sorted Array [binary search] <c++> 给出排序好的一维无重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数的下标,如果没有,返回-1.时间复杂度限制\(O(log_2n)\) C++ 我的想法是先找到数组中最大值的位置.然后以此位置将数组一分为二,然后在左右两部分分别寻找target. 二分寻找最大值的时候,因为左半部分的数一定大于nums[l],所以n…
Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1. Example 1: Input: nums = [-1,0,3,5,9,12], target = 9 Out…
This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates. 思路 该题是[leetcode]33. Search in Rotated Sorted Array旋转过有序数组里找目标值 的followup 唯一区别是加了line24-26的else语句来skip duplicates 代码 class Solution { public boolean sear…
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 解题思路: 参考Java for LeetCode 033 Search in Rota…