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.
You may assume no duplicate exists in the array.
题意：在旋转的已排序数组中找出特定值的索引
思路：1.用二分法找出数组中最小值的索引
2.用最小值把数组分为两部分，确定特定值在哪个部分，然后对单独的一个部分利用二分法
（感觉这个思路比《剑指OFFER》上面的那个思路容易理解）
1 int search(int* nums, int numsSize, int target) { 2 int lo=0,hi=numsSize-1,mid; 3 while(lo<hi){ 4 mid=(lo+hi)/2; 5 if(nums[mid]>nums[hi]) 6 lo=mid+1; 7 else 8 hi=mid; 9 } 10 if(nums[numsSize-1]>=target) 11 hi=numsSize-1; 12 else{ 13 hi=lo-1; 14 lo=0; 15 } 16 while(lo<=hi){ 17 mid=(lo+hi)/2; 18 if(target>nums[mid]) lo=mid+1; 19 else if(target<nums[mid]) hi=mid-1; 20 else return mid; 21 } 22 return -1; 23 }
另外，讨论区的另外两种做法让人佩服死了。。。膜拜一下