33. Search in Rotated Sorted Array

binary search O(logn)
先判断左半边数多还是右半边数多，再对每一种情况分类讨论

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1
        while (l <= r):
            m = l + (r - l) // 2
            if nums[m] == target:
                return m
            elif nums[m] > nums[r]:
                if nums[l] <= target < nums[m]:
                    r = m - 1
                else:
                    l = m + 1
            else:
                if nums[m] < target <= nums[r]:
                    l = m + 1
                else:
                    r = m - 1
        return -1

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        int l = 0, r = n - 1;
        while (l <= r) { // 这道题是要找数，所以要=
            int mid = (l + r) / 2;
            if (target == nums[mid]) { // 找到直接返回
                return mid;
            } else if (nums[mid] > nums[r]) {
                if (nums[l] <= target && target < nums[mid]) { // 检查是否在单调区间
                    r = mid - 1;
                } else {
                    l = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[r]) {
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
        }
        return -1;
    }
};