34. 在排序数组中查找元素的第一个和最后一个位置 - 力扣（LeetCode）
根据二分函数，得到>=target和<=target的两个，分别是答案的l和r
class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {if (nums.size() == 0) return { -1, -1 };vector<int> ans(2);int l = 0, r = nums.size() - 1;while (l < r) {int mid = (l + r + 1) / 2;if (nums[mid] <= target) l = mid;else r = mid - 1;}if (nums[l] != target) return { -1, -1 };ans[1] = l;l = 0, r = nums.size() - 1;while (l < r) {int mid = (l + r) / 2;if (nums[mid] >= target) r = mid;else l = mid + 1;}ans[0] = l;return ans;}
};
33. 搜索旋转排序数组 - 力扣（LeetCode）
先确定两个有序区间的分界点，然后分别对两个区间进行二分
需要考虑数组长度为2，且旋转过的情况
class Solution {
public:int search(vector<int>& nums, int target) {int l = 0, r = nums.size() - 1;int t = 0;if (*nums.begin() > *nums.rbegin()) {while (l < r) {int mid = (l + r) / 2;if (mid - 1 >= 0 && nums[mid] < nums[mid - 1]) {t = mid;break;}else if (mid + 1 < nums.size() && nums[mid] > nums[mid + 1]) {t = mid + 1;break;}else if (nums[mid] > nums[0]) l = mid;else if (nums[mid] < nums[0]) r = mid;}}if (t) {auto it = lower_bound(nums.begin(), nums.begin() + t, target);if (it != nums.end() && *it == target) return it - nums.begin();it = lower_bound(nums.begin() + t, nums.end(), target);if (it != nums.end() && *it == target) return it - nums.begin();else return -1;}else {auto it = lower_bound(nums.begin(), nums.end(), target);if (it != nums.end() && *it == target) return it - nums.begin();else return -1;}}
};
