退化版问题二: Find Minimum in Rotated Sorted Array 在回到最开始的问题之前,我们再来看一个比刚才稍微进化一点(但依然是原题退化版)的问题。问题描述如下: Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.(i.e.,
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...
Suppose a sorted array 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 exists in the array...
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. 题解: 这道题是一道常见的二分查找法的变体题。 要解决这道题,需要明确rotated sorted array的特性,那么就是至少有一侧是排好序的(无论pivot...
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. ...
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. ...
This is a follow up problem toSearch in Rotated Sorted Array, wherenumsmay contain duplicates. Would this affect the run-time complexity? How and why? 题目大意 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 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。
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。
https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ 这两道题都是属于循环有序数组的查找问题,无论是查找具体元素的下标还是检查一个目标值是否在数组中,主要的判断逻辑都是一致的。 思路 这种将有序数组翻转成循环有序数组的解决办法是将原数组分段。用首元素start、中间元素mid、尾元素end,可以...