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,
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 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 duplicate exists in the array...
退化版问题二: Find Minimum in Rotated Sorted Array 在回到最开始的问题之前,我们再来看一个比刚才稍微进化一点(但依然是原题退化版)的问题。问题描述如下: 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 ...
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...
Can you solve this real interview question? Search in Rotated Sorted Array - There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1
Lintcode62 Search in Rotated Sorted Array solution 题解 【题目描述】 Suppose a sorted array 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...
You may assume no duplicate exists in the array. 就是在一个旋转后的数组中寻找target数据。 很明显,最简单的方法就是遍历一次,不过题目的意思明显不是要求这么做的,主要考察的还是使用二分查找,所以这里使用二分查找来做。 代码如下: public class Solution { ...
1. Description Search in Rotated Sorted Array II 2. Solution class Solution{public:boolsearch(vector<int>&nums,inttarget){intsize=nums.size();intleft=0;intright=size-1;while(left<=right){intmid=(left+right)/2;if(nums[mid]==target){returntrue;}if(nums[left]==nums[mid]){left++;cont...
classSolution{public:intsearch(vector<int>&nums,inttarget){if(nums.empty())return-1;intstart=0;intend=nums.size()-1;while(start+1<end){intmid=start+(end-start)/2;if(target==nums[mid])returnmid;if(nums[start]<nums[mid]){//可判断start 到 mid 是有序的。if(target>=nums[start]&&ta...
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....