153. Find Minimum in Rotated Sorted Array(二分法)
題目描述:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
二分法,如果某一半是單增的,那麼最小值一定在另一半或者這一半的頭位置。
class Solution { public: int findMin(vector<int> &nums) { int head=0; int tail=nums.size()-1; if (nums[head]<nums[tail]) return nums[head]; while(head<tail) { int mid=(head+tail)/2; if (nums[mid]<nums[tail]) tail=mid; else head=mid+1; } return nums[head]; } };
相關推薦
153. Find Minimum in Rotated Sorted Array(二分法)
題目描述:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ 二分法,如果某一半是單增的,那麼最小值一定在另一半或者這一半的頭位置。 class Solution { public: int fi
LeetCode 153 Find Minimum in Rotated Sorted Array
== pad assume mat rac wid san orm trac Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5
153. Find Minimum in Rotated Sorted Array
nim OS duplicate you put clas 一半 amp pos 問題描述: Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforeh
153. Find Minimum in Rotated Sorted Array - LeetCode
etc mini https problem 如果 .com nim -i ref Question 153.?Find Minimum in Rotated Sorted Array Solution 題目大意:給一個按增序排列的數組,其中有一段錯位了[1,2,3,4,
Leetcode|Find Minimum in Rotated Sorted Array[二分查詢]
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). Find the minimu
#Leetcode# 153. Find Minimum in Rotated Sorted Array
exists num duplicate empty amp break NPU 一個 play https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ Suppose an array s
153. Find Minimum in Rotated Sorted Array - Medium
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
Leetcode 153. Find Minimum in Rotated Sorted Array
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 &nbs
【LeetCode-面試演算法經典-Java實現】【153-Find Minimum in Rotated Sorted Array(找旋轉陣列中的最小數字)】
原題 Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0
[LeetCode]153.Find Minimum in Rotated Sorted Array
【題目】 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). Fi
LeetCode | 153. Find Minimum in Rotated Sorted Array
題目: 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 becom
LeetCode 153. Find Minimum in Rotated Sorted Array(旋轉陣列查詢)
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). Find t
【LeetCode & 劍指offer刷題】查詢與排序題7:11旋轉陣列的最小數字(153. Find Minimum in Rotated Sorted Array)(系列)
【LeetCode & 劍指offer 刷題筆記】目錄(持續更新中...) 153. Find Minimum in Rotated Sorted Array Suppose an array sorted in ascending ord
153. Find Minimum in Rotated Sorted Array(旋轉數組的最小數字)(leetcode)
sorted order 最小數 技術 變種 and 分法 目標 asc Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e
新手算法學習之路----二分法Find Minimum in Rotated Sorted Array
有一個 序列 思路 pan ron write -1 需要 cnblogs 題目:假設一個旋轉排序的數組其起始位置是未知的(比如0 1 2 4 5 6 7 可能變成是4 5 6 7 0 1 2)。 你需要找到其中最小的元素。 你可以
LeetCode 153, 154. Find Minimum in Rotated Sorted Array I & II
153. Find Minimum in Rotated Sorted Array 二分題目,由於rotated存在,a[mid]<key不能判斷在哪一邊搜尋。 可以根據a[low]與a[high]的關係,來判斷哪一邊有序,哪一邊存在rotate,進而縮小搜尋區間。 開區間寫法:(
Leetcode|Find Minimum in Rotated Sorted Array II(有重複元素的二分查詢)
Follow up for “Find Minimum in Rotated Sorted Array”: What if duplicates are allowed? Would this affect the run-time complexity? How and why?
LeetCode--153 尋找旋轉排序陣列中的最小值 (Find Minimum in Rotated Sorted Array)(C語言版)
題目描述: 解題思路:可以使用順序遍歷,將最小值記錄下來,時間複雜度為O(n),但是根據題目描述,該陣列為旋轉排序陣列,這裡就可以聯想到二分查詢,試著用二分查詢的思想,首先使用Lindex和Rindex記錄左右區間的下標,然後再用Mindex表示區間中間元素的下標,根據
leetcode解題之153&154. Find Minimum in Rotated Sorted Array版(在旋轉的陣列中查詢最小數字)
153.Find Minimum in Rotated Sorted Array Suppose an array sorted in ascending order is rotated
Find Minimum in Rotated Sorted Array II
number mce minimum div ant remove span col mean The worst situation O(N). Actually we can either just loop through, or we can compare nu