与Leetcode Find Minimum in Rotated Sorted ArrayLeetcode Find Minimum in Rotated Sorted Array II是一个类型的题目。 用binary search,Complexity: O(logn), 关键点在于每次要能判断出target位于左半还是右半序列。解这题得先在纸上写几个rotated sorted array的例
[LeetCode]33.Search in Rotated Sorted Array ;后半部分为严格递增数组。【代码】 【分析二】 对于一个数组4,5,6,7,0,1,2 你首先找到那个转折点,就是大于下一个相邻数字的那个数字的下标,在这个数组就是数字7的下标3。 步骤: 1 找到转折点下标,把数组分成两个有序的子数组 2 如果转折点下标返回-1...
改进如下(改进如下(参考leetcode某大神的解法)) publicintsearchInRotatedSortedArray(int[] nums,inttarget){intlow=0;inthigh=nums.length-1;while(low < high){intmid=low +((high-low) >>2);if((nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid])) low = mid+1;elseh...
33. Search in Rotated Sorted Array # 题目# 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 become [4,5,6,7,0,1,2]). You are given a target value to search. If found in the array return its ...
【LeetCode 33】Search in Rotated Sorted Array 题意: 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 foun......
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. 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,...
【LeetCode】154. Find Minimum in Rotated Sorted Array II (cpp) 题目和测试用例略233 本题作为153题的升级版,难点在于数组中的数可能有重复的。 这种问题很容易想到用二分搜索的思想来解决(毕竟如果暴力求解时间复杂度只有O(N),比O(N)小的就是O(logN)了233)。设l是考虑的左边界,r是右边界,m是计算...
LeetCode-33. Search 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 [4,5,6,7,0,1,2])....
LeetCode 33. Search in Rotated Sorted Array 题目描述(中等难度) 开始的时候想复杂了,其实就是一个排序好的数组,把前边的若干的个数,一起移动到末尾就行了。然后在 log (n) 下找到给定数字的下标。 总的来说,log(n),我们肯定得用二分的方法了。
81. Search in Rotated Sorted Array II 如果数组中有重复元素。参见https://leetcode.com/problems/search-in-rotated-sorted-array-ii/discuss/28218/My-8ms-C++-solution-(o(logn)-on-average-o(n)-worst-case)。 class Solution { public: bool search(vector<int>& nums, int target) { int r = ...