旋转数组二分查找法搜索

题目来源leetcode 33.Search in Rotated Sorted Array 地址：点击打开链接

代码

class Solution {public:    int search(vector<int>& nums, int target) {        int n = nums.size();    int first = 0;    int last = n - 1;    int mid = first + (last - first) / 2;    if (nums[mid] == target)  return mid;    while (first != last) {if (nums[first]<=nums[mid]) {//左侧有序if (target<nums[mid] && nums[first] <= target)last = mid - 1;elsefirst = mid + 1;mid = first + (last - first) / 2;if (nums[mid] == target)return mid;}else if(nums[mid]<=nums[last]) {//右侧有序if (nums[mid]<target&&target <= nums[last])first = mid + 1;elselast = mid-1;mid = first + (last - first) / 2;if (nums[mid] == target)return mid;}}    return -1;    }};

笔记：这道题用二分法。首先思路上要明晰：中值左右两边至少有一边是有序的，所以分为两种情形，见注释。即便如此，在测试case时还是多次出现错误结果，改的心烦意乱，还用了本地调试器帮忙。上面的答案跟许多大神的相比不够简洁，跟坨shit一样，真不知道怎么才可写出简洁准确的代码。新手起步，挫折感较强。