本文共 1009 字，大约阅读时间需要 3 分钟。

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].

class Solution {public:    vector
   
     searchRange(int A[], int n, int target) {        vector
    
      re(2, -1);        if(n == 0) return re;        int low = 0;         int high = n - 1;        while(low <= high)        {            int mid = (low + high) >> 1;            if(A[mid] == target)            {                low = mid;                while(low - 1>= 0 && A[low - 1] == target) --low;                while(mid + 1 <= high && A[mid + 1] == target) ++mid;                re[0] = low;                re[1] = mid;                return re;            }            else if(A[mid] < target) low = mid + 1;            else high = mid - 1;        }        return re;    }};
    
   

转载地址：http://relji.baihongyu.com/