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# 34_Search_for_a_Range

Posted on 2016-08-14 Edited on 2020-01-12 In leetcode Views: Views: Disqus:

## question

### 34. Search for a Range

Original Page

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]`.

## analysis

一开始就想到了二分查找，但是原来做二分查找的时候一般都是找到确定的那个数就完成了，
这里的情况比较特殊，需要找到整个区间，所以需要两遍查找，并且一个是找到小于target
的最大索引，一个是找到大于target的最大索引，代码参考leetcode discuss,这位仁
兄也做了详细的分析解释。

## code

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    class Solution {
    public:
        vector<int> searchRange(vector<int>& nums, int target) {
            vector<int> ret(2, -1);
            int i = 0, j = nums.size() - 1;
            int mid;
            while(i < j){
                mid = (i + j) / 2;
                if(nums[mid] < target) i = mid + 1;
                else j = mid;
            }
            if(nums[i] != target) return ret;
            else {
                ret[0] = i;
                if((i+1) < (nums.size() - 1) && nums[i+1] > target){
                    ret[1] = i;
                    return ret;
                }
            }   //一点小优化
            j = nums.size() - 1;
            while(i < j){
                mid = (i + j) / 2 + 1;
                if(nums[mid] > target) j = mid - 1;
                else i = mid;
            }
            ret[1] = j;
            return ret;
        }
    };

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- Post author: Nicksxs
- Post link: https://nicksxs.me/2016/08/14/34-Search-for-a-Range/
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