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
You may assume no duplicate exists in the array. Tag: Array; Binary Search 体会: 这道题和Find the Minimum In Rotated Sorted Array肯定是有联系啦,毕竟给的都是Rotated Sorted Array这种结构。思路是:从middle处开始左右两部分,一部分是sorted的了,另一部分rotated或者sorted。我们要去分析sorted的那个部分。
publicintsearch(int[]nums,inttarget){intstart=0;intend=nums.length-1;//找出最小值的数组下标/* while (start < end) {int mid = (start + end) / 2;if (nums[mid] > nums[end]) {start = mid + 1 ;} else {end = mid;}}int bias = start;*///找出最大值的数组下标while(start<end...
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. If found in the array return its index, otherwise return -1. You may assume no...
题目地址:https://leetcode.com/problems/search-in-rotated-sorted-array/description/ 题目描述 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]). ...
Suppose a sorted array is rotated at some pivot unknown to you beforehand. 2.1 数组 5 (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. If found in the array return its index, otherwise return -1. ...
sort that is theoretically optimal in terms of the total number of writes to the original array, unlike any other in-place sorting algorithm. It is based on the idea that the permutation to be sorted can be factored into cycles, which can individually be rotated to give a sorted result....
33. Search in Rotated Sorted Array 寻找旋转排序数组 There is an integer array nums sorted in ascending order (with distinct values). 一个升序排序的整数数组 Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that th...
题目33. Search in Rotated Sorted Array 题解 题目意思就是在一个上升序列中截断成AB 两个子序列 然后BA拼接成一个新的序列, 而题目要求...
// Search in Rotated Sorted Array II// Time Complexity: O(n),Space Complexity: O(1)classSolution{public:boolsearch(constvector<int>&nums,inttarget){intfirst=0,last=nums.size();while(first!=last){constintmid=first+(last-first)/2;if(nums[mid]==target)returntrue;if(nums[first]<nums[...