LeetCode 34 Search for a Range

Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

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

解题思路：有了上一题的启发，这一题的解决办法是利用二分法处理，利用无递归的二分法算法实现题目要求的O(log n)的要求，然后利用前后遍历实现找到范围所在。但也要考虑到，当重复项过多，该算法会退化到O(n)的复杂度。

代码如下：

 public int[] searchRange(int[] nums, int target) {       int len = nums.length;       int l =0; int r = len - 1;       int mid =(l+r)/2;       int i=0,j=0;       int []re = {-1,-1};       while(l<=r){    if(target<nums[l]||target>nums[r])    return re;       if(nums[mid]==target){       i = mid;       j = mid;       if(mid>0){    int t = i;    while(nums[--t]==target){    i=t;    if(i>0){    continue;    }else{    break;    }    }    }       if(mid<len-1){       int g =j;   while(nums[++g]==target){   j =g;       if(j<len-1){       continue;       }else{       break;       }       }   }       int []re1 = {i,j};       return re1;       }else if(nums[mid]<target){       l = mid+1;       }else{       r =mid-1;       }      mid =(l+r)/2;       }  return re;    }