思路:先二分查找到一个和target相同的元素,然后再左边二分查找左边界,右边二分查找有边界。
class Solution { public: int begin = -1,end = -1; int ends; int lSearch(int left,int right,vector<int>& nums,int target) { if(left > right) return -1; int mid = (left + right) / 2; if(nums[mid] == target){ if(mid == 0 || (mid > 0 && nums[mid - 1] < target)) return mid; else return lSearch(left,mid - 1,nums,target); }else return lSearch(mid + 1,right,target); return -1; } int rSearch(int left,int target) { if(left > right) return -1; int mid = (left + right) / 2; if(nums[mid] == target){ if(mid == ends || (mid < ends && nums[mid + 1] > target)) return mid; else return rSearch(mid + 1,target); }else return rSearch(left,target); return -1; } int midSearch(int left,int target) { if(left > right) return -1; int mid = (left + right) / 2; if(nums[mid] == target){ return mid; } else if(nums[mid] < target) return midSearch(mid + 1,target); else if(nums[mid] > target) return midSearch(left,target); return -1; } vector<int> searchRange(vector<int>& nums,int target) { ends = nums.size() - 1; int mid = midSearch(0,ends,target); if(mid != -1){ begin = lSearch(0,mid,target); end = rSearch(mid,target); } vector<int> ans; ans.push_back(begin); ans.push_back(end); return ans; } };