1)如果nums[mid] < target <= nums[right],说明target在右边区间里,则left = mid + 1; 2)否则在左边区间里,搜索左边区间,right = mid - 1; 3. nums[mid] > nums[right],说明[elft, mid]区间是在左边的递增区间,然后判断target是否在这个左边区间里 1)如果nums[left] <= target < nums[mid],说...
Input:nums = [2,5,6,0,0,1,2], target = 3Output:false Constraints: 1 <= nums.length <= 5000 -104<= nums[i] <= 104 numsis guaranteed to be rotated at some pivot. -104<= target <= 104 Follow up:This problem is similar toSearch in Rotated Sorted Array, butnumsmay containdup...
初始版本 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). You are given a target value to search. If found in the array return its index, otherwise return -1. You ma...
题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ 题目: Follow up for “Search in Rotated Sorted Array”: What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. ...
Write a function to determine if a given target is in the array. The array may contain duplicates. 这道题很简单,直接遍历即可。也可以使用二分查找。 代码如下: public class Solution { /* * 最简单的就是遍历 * */ public boolean search(int[] nums, int target) ...
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 ...
This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 描述 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] )。
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. ...
[medium] > target)high = medium - 1; else low = medium+1; } return -1; } int search(int* nums, int numsSize, int target) { int low = 0; int high = numsSize - 1; while(low <= high) { if(nums[low] <= nums[high]) { int re = bs(nums+low,high - low + 1,target)...
}else{if(A[left] <= target && A[mid] > target) right = mid -1;elseleft = mid +1; } }return-1; } }; 本文转自博客园Grandyang的博客,原文链接:在旋转有序数组中搜索[LeetCode] Search in Rotated Sorted Array