Search for a Range

Online to see better ideas, Online said is, first use the dichotomy to find the left point, and then use a binary search to find the right breakpoint. The problem is solved.

My thinking is not very good, I was the first to find the index equal to target, and then the center to the two sides of the expansion.

classSolution { Public: Vector<int> Searchrange (vector<int>& Nums,inttarget) {        intlen=nums.size (); intL=0, r=len-1; intmid; Vector<int>Res; intflag=0;  while(l<=r) {Mid=l+ (r-l)/2; if(nums[mid]<target) L=mid+1; Else if(nums[mid]>target) R=mid-1; Else{flag=1;  Break; }        }        if(flag==0) {Res.push_back (-1); Res.push_back (-1); }        Else{L=mid; R=mid;  while(l>=0&&nums[l]==target) {                if(nums[l]==target) L--; }             while(r<=len-1&&nums[r]==target) {                if(nums[r]==target) R++; } res.push_back (L+1); Res.push_back (R-1); }       returnRes; }};

Search for a Range