【LeetCode算法练习（C++）】Search in Rotated Sorted Array

题目：
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.

链接：Search in Rotated Sorted Array
解法：一次二分查找确定两个连续递增序列的左右边界，判断待查找数字处于哪个序列中，再在该序列中进行一次二分查找找到位置。时间O(logn)

class Solution {public:    int search(vector<int>& nums, int target) {        if (nums.size() == 0) return -1;        else if (nums.size() == 1) {            if (nums[0] == target) return 0;            else return -1;        }        int l = 0, r = nums.size() - 1;        while (l < r && nums[l] > nums[r]) {            int m = (l + r) / 2;            if (nums[m] > nums[l]) l = m;            else if (nums[m] < nums[r]) r = m;            else break;        }        if (target >= nums[0] && nums[l] > nums[r]) {            r = l;            l = 0;        } else if (nums[l] > nums[r]) {            l = l + 1;            r = nums.size() - 1;        }        if (nums[l] > target || nums[r] < target) return -1;        while (l <= r) {            int m = (l + r) / 2;            if (nums[m] == target) return m;            else if (nums[m + 1] == target) return m + 1;            else if (nums[m] < target && nums[m + 1] > target) return -1;            else if (nums[m] < target) l = m;            else r = m;        }    }};           

Runtime: 6 ms