leetcode 033 Search in Rotated Sorted Array
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题目
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
).
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.
思路:
题目意思是:有一个已经排好序的数组,但是这个数组从中间某一个点旋转了一下。给一个数,查找它在这个数组中的位置。
我的算法是:先找出pivot所在位置,然后然后比较数组第一个数和target的大小,若nums[0] <= target,则在0至piovt+1之间用二分查找;否则在pivot+1至最后用二分查找。
说实话,我不明白这题目为什么会是hard级别的,就用最简单的直接查找也能在1ms完成。
代码:
public int search(int[] nums, int target){ int pivot = 0; while(pivot < nums.length - 1 && nums[pivot] < nums[pivot + 1]) pivot++; int res; if(nums[0] <= target) res = Arrays.binarySearch(nums,0,pivot+1,target); else res = Arrays.binarySearch(nums,pivot+1,nums.length,target); return res < 0 ? -1 : res;}
结果细节(图):
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