LeetCode 33 Search in Rotated Sorted Array [binary search] 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$O(log_2n)$
KEY WORDS: [Rotated sorted array] [find target] [contain duplicates] publicbooleansearch(int[] nums,inttarget) {if(nums ==null|| nums.length == 0)returnfalse;intleft = 0;intright = nums.length - 1;while(left <=right) {intmid = left + (right - left) / 2;if(nums[mid] ==targ...
Search in rotated sorted array - Leetcode 33 - Python 呼吸的chou 0 0 Code-It-Yourself! Tetris - Programming from Scratch (Quick and Simple C++) 呼吸的chou 4 0 Decode Ways - Dynamic Programming - Leetcode 91 - Python 呼吸的chou 0 0 ...
33. Search in Rotated Sorted Array 题解 528. Random Pick with Weight 题解 按值范围折半查找 折半查找还可以应用于非有序区间查找满足特定条件的值。该场景下所找的值在已知范围内,这时折半的不是索引,而是值本身所在的范围。算法基本框架如下: //287. Find the Duplicate Number int findDuplicate(vecto...
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进阶习题-题目选自leetcode-33.Search in Rotated Sorted Array 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] )。 搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 。
81 Search in Rotated Sorted Array II 详细解答 81Searchin Rotated Sorted Array II 详细解答 由于题目要求时间复杂度是O(logn), 很容易想到二分法。 但应该怎么构造二分? 举例:nums...[l],说明从 l – > mid是个递增的序列,再寻找target是否在(nums[l],nums[3]) 之间,如果是,缩小范围使得 r = mid...
Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 数组的中点肯定就是平衡BST的根节点,以此规律递归。 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; ...
Binary-Search-1 Problem1 Search a 2D Matrix(https://leetcode.com/problems/search-a-2d-matrix/) Problem2 Search in a Rotated Sorted Array (https://leetcode.com/problems/search-in-rotated-sorted-array/) Problem3 Search in Infinite sorted array: https://leetcode.com/problems/search-in-a-...
33. Search in Rotated Sorted Array https://leetcode.com/problems/search-in-rotated-sorted-array/description/ classSolution {public:intsearch(vector<int>& nums,inttarget) {if(nums.size() ==0)return-1;inti =0, j = nums.size()-1;while(i +1<j) {intm = i + (j - i) /2;if(...