leetcode--34. 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].

题目大意：给定一个排序的数组和一个目标数，寻找该目标数在数组中出现的首尾位置。如题中例子，8在数组中是第3、4个。

思路：排序数组->二分查找。如果nums[mid] > target 那么end需要变为mid-1；如果nums[mid] < target，那么begin需要变为mid+1；如果nums[mid]==target，那么首尾值肯定是在左右两边，注意边界处理。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int begin = 0;        int end = nums.size()-1;        vector<int> result;                while(nums[begin] <= nums[end]){           int mid = (begin+end)/2;            if(nums[mid] == target){                begin = mid ;                 while(nums[begin] == target && begin>=0){                    --begin;                }                result.push_back(begin+1);                end = mid;                while(nums[end] == target && end <= nums.size()-1){                    ++end;                }                result.push_back(end-1);                return result;            }            else if(nums[mid] < target){                begin = mid + 1;            }            else if(nums[mid] > target){                end = mid -1;            }        }        result.push_back(-1);        result.push_back(-1);        return result;    }};

leetcode编译通过。