不过要注意没有pivot的特殊情况,即array是sorted但没有被rotated。 在得到pivot后,我们便可以在pivot划分的两个区域内分别用binary search搜索target。时间复杂度为O(logn)。 1classSolution {2public:3intsearch(intA[],intn,inttarget) {4intpivot = search_pivot(A,0,n-1);5if(pivot == -1)returnbinary...
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. 做完了Find Minimum in Rotated Sorted Array之后,对这题又发现了一种非常简单的做法。 Find Minimum in Rotated Sorted Array的思路如下: 当nums...
publicintsearch(int[]nums,inttarget){intlo=0,hi=nums.length-1;while(lo<=hi){intmid=lo+(hi-lo)/2;intnum=nums[mid];//nums [ mid ] 和 target 在同一段if((nums[mid]<nums[0])==(target<nums[0])){num=nums[mid];//nums [ mid ] 和 target 不在同一段,同时还要考虑下变成 -inf 还...
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...
Search in Rotated Sorted Array https://leetcode.com/problems/search-in-rotated-sorted-array/description/ Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums. ...
与情况2的给定数组nums一样,不过情况3查找的数不在Subnums里边,那么,就让low =medium + 1,high不变,重新开始判断这个数组。 情况4. nums = {4,5,6,7,8,1,2},target = 8 与情况3需要的做法是一样的,如果在Subnums = {4,5,6,7}里边没有查到想要的,那么low = medium +1,high不变,重新从情况...
33. Search in Rotated Sorted Array 题解 题目意思就是在一个上升序列中截断成AB 两个子序列 然后BA拼接成一个新的序列, 而题目要求 我们查找在这个序列中是否存在target,如果看到这里就会发现 这个很简单啊,查找一个序列中是否有某个数字 直接遍历就好了 方法木佬佬,但是 最重要的是题目末尾有一句 ...
这道题一开始看半天没有弄明白后来恍然大悟直接看注释基于常规二分查找算法重点在于 我们确定lr边界变化的时候需要考虑mid所在不同分段存在不同情况 还有一种解法 找到反转的in...
sort that is theoretically optimal in terms of the total number of writes to the original array, unlike any other in-place sorting algorithm. It is based on the idea that the permutation to be sorted can be factored into cycles, which can individually be rotated to give a sorted result....
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. ...