二分法相关 153. Find Minimum in Rotated Sorted Array 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 minimum element. You may assume no duplicate exists in the array. (M…

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 minimum element. You may assume no duplicate exists in the array. 不知道这道题为什么难度是Medium,感觉蛮简单的. 仅仅须要找到第一个大于它后面的数,它后面的数就…

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] ). 请找出其中最小的元素. 你可以假设数组中不存在重复元素. 示例 1: 输入: [3,4,5,1,2] 输出: 1 示例 2: 输入: [4,5,6,7,0,1,2] 输出: 0 最好用high值来判断 class Solution { public: int findMin(vector<int>& nums) { int len =…

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be…

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 minimum element. You may assume no duplicate exists in the array. 这道寻找旋转有序数组的最小值肯定不能通过直接遍历整个数组来寻找,这个方法过于简单粗暴,这样的话,旋不…

题目概述: 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).Find the minimum element.You may assume no duplicate exists in the array. 解题思路: 在一个有顺序的数组中查找个最小值,很容易想到的就是二分法,的确,这里用的就是二分,…

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 minimum element. You may assume no duplicate exists in the array. 二分查找就行了. class Solution { public: int findMin(vect…

原题链接在这里:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ 题目: Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted arra…

原题链接在这里:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ Method 1 就是找到第一个违反升序的值,就是最小值,若是没有,那么第一个值就是最小值. Time O(n). Method 2 Binary Search. 与Find Peak Element类似. while 的条件是 l < r 并且 nums[l] > nums[r], 不然 l 到 r 这一段就是sorted的. nums[mi…

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be…