Suppose an array sorted in ascending order 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 ...
classSolution {publicintsearch(int[] nums,inttarget) {returnsearch(nums, 0, nums.length - 1, target); }privateintsearch(int[] nums,intlow,inthigh,inttarget) {if(low >high)return-1;intmid = (low + high) / 2;if(nums[mid] ==target)returnmid;if(nums[mid] <nums[high]) { //后...
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...
Search in Rotated Sorted Array I 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, otherwise return -1. You may ...
Search in Rotated Sorted Array I && II Leetcode 对有序数组进行二分查找(下面仅以非递减数组为例): 1. int binarySort(int A[], int lo, int hi, int target) 2. { 3. while(lo <= hi) 4. { 5. int mid = lo + (hi - lo)/2; ...
This is a follow up problem toSearch in Rotated Sorted Array, wherenumsmay contain duplicates. Would this affect the run-time complexity? How and why? 在旋转有序数组中搜索二。 这题跟33题求的一样,多一个条件是input里面有重复数字。依然是用二分法做,但是worst case很可能会到O(n);而且其中一开...
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 。
target小于nums[0] && mid 大于 target也有两种情况 Runtime: 0 ms, faster than 100.00% of Java online submissions for Search Insert Position. \classSolution{publicintsearch(int[]nums,inttarget){intn=nums.length;if(n==0)return-1;intleft=0;intright=n-1;while(left+1<right){intmid=left+(...
image.png 二分查找的扩展 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。
(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. 原问题链接:https://leetcode.com/problems/search-in-rotated-sorted-array/...