在旋转排序数组中查找元素(Search in Rotated Sorted Array) lintcode:题号——62,难度——medium 2 描述 给定一个有序数组,但是数组以某个元素作为支点进行了旋转(比如,0 1 2 4 5 6 7 可能成为4 5 6 7 0 1 2)。给定一个目标值target进行搜索,如果在数组中找到目标值返回数组中的索引...
空间复杂度O(1)#include<iostream>usingnamespacestd;classSolution{public:intSearchRotatedSortedArray(intA[],intn,inttarget){intstart =0;intend = n;intmiddle = start + (end - start) /2;while(start != end) {if(target == A[middle])returnmiddle;if(A[start] < A[middle]) {if((target ...
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...
Description: 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 ma...
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]). ...
Search in Rotated Sorted Array 描述 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 ...
This algorithm is a combination of radix sort and quicksort. Pick an element from the array (the pivot) and consider the first character (key) of the string (multikey). Partition the remaining elements into three sets: those whose corresponding character is less than, equal to, and greater...
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 我们查找在这个序列中是否存在target,如果看到这里就会发现 这个很简单啊,查找一个序列中是否有某个数字 直接遍历就好了 方法木佬佬,但是 最重要的是题目末尾有一句 ...
LeetCode-33-搜索旋转排序数组(Search in Rotated Sorted Array)33. 搜索旋转排序数组整数数组 nums 按升序排列,数组中的值 互不相同。在传递给函数之前,nums 在预先未知的某个下标 k(0 <= k < nums.length)上进行了 旋转,使数组变为 [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1...
Search in Rotated Sorted Array II Follow up for "Search in Rotated Sorted Array": What ifduplicatesare>[1,3,1,1,1], 3 Would this affect the run-time complexity? How and why? --run-time的最坏情况是O(n)了,因为特殊情况出现的时候需要额外处理,可能做线性搜索 ...