Leetcode34--->search for a range (to find out what the given value appears in the sorted array)

title: Given a sorted array, find out the scope of the given target value, the complexity of the algorithm is required in O (Logn), and if not found, returns [-1,-1];

Example:

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
Return [3, 4] .

Problem Solving Ideas:

When we see the time complexity O (logn), we naturally want to use the two-point method, this problem, we naturally also use a two-point method;

Mid = (start + end)/2; There are three relationships between Target and Nums[mid]:

1) Target <nums[mid], then the latter part is discarded, end = mid-1;

2) Target > Nums[mid], the first half is discarded, start = mid + 1;

3) target = Nums[mid], this situation can not simply discard which half, but to go from the middle to the ends of the traversal, until the boundary is found;

The code is as follows:

1  Public classSolution {2      Public int[] Searchrange (int[] Nums,inttarget) {3         if(Nums = =NULL|| Nums.length < 1)4             return New int[]{-1,-1};5         int[] res =New int[2];6Res[0] = 1;7RES[1] = 1;8         intStart = 0;9         intEnd = Nums.length-1;Ten         intMID = 0; One          while(Start <=end) A         { -Mid = (start + end)/2; -             if(Nums[mid] <target) theStart = mid + 1; -             Else if(Nums[mid] >target) -End = Mid-1; -             Else +             { -Start =mid; +End =mid; A                  while(Start >= 0 && Nums[start] = =target) atstart--; -                  while(End < Nums.length && Nums[end] = =target) -end++; -                 return New int[]{start + 1, end-1}; -             } -         } in         returnRes; -     } to}

Leetcode34--->search for a range (to find out what the given value appears in the sorted array)