描述 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.
分析 二分查找，难度主要在于左右边界的确定。
代码
class Solution {public: int search(vector<int>& nums, int target) { int first = 0; int last = nums.size(); while (first != last) { int mid = (first + last) / 2; if (nums[mid] == target) return mid; if (nums[first] <= nums[mid]) { if (nums[first] <= target && target < nums[mid]) last = mid; else first = mid + 1; } else { if (nums[mid] < target && target <= nums[last - 1]) first = mid + 1; else last = mid; } } return -1; }};新闻热点
疑难解答
