调试了半天终于过了……
神人题目,主要是情况太太太多了,有先找到左边界的情况、先找到右边界的情况、找到中间节点之后要依次找左右边界的情况……其实要是弄多一点循环应该就不会像我写的这么复杂,但我太懒了就是不想多开循环。
class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {int left=0;int right=nums.size()-1;vector<int> result{-1,-1};int recordl=left;int recordr=right;while(left<=right){int mid=(left+right+1)/2;if(nums[mid]>target) right=mid-1;else if(nums[mid]<target) left=mid+1;else{if(mid+1==nums.size()||nums[mid+1]!=target){result[1]=mid;right=mid-1;}if(mid==0||nums[mid-1]!=target){result[0]=mid;left=mid+1;}if(result[1]!=-1&&result[0]!=-1) return result;else if(result[0]!=-1&&recordl!=0){left=recordl;right=recordr;recordl=0;}else if(result[0]==-1&&recordl==0){recordl=mid+1;recordr=right;right=mid-1;}else if(result[0]==-1) right=mid-1;else left=mid+1;}}return result;}
};