1、问题描述
Given an array of integers sorted in ascending order, 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].
给定一个按照上升排序的数组和一个数，找出这个数在该数组中的开始位置和结束位置。如果该数不存在数组中。返回[-1,-1].
2、问题分析
题目要求在log(N)的时间内。选择二分查找，现在数组中找出该数的一个位置，然后向左右两边扫描，找出左边的位置和右边的位置。
3、代码
vector<int> searchRange(vector<int>& nums, int target) {
// 数组是有序的 ，先用二分查找找出一个目标数。然后向两边寻找
vector<int> ans;
if(nums.size() == )
{
ans.push_back(-);
ans.push_back(-);
return ans;
}
int left = ,right = nums.size()-;
int index = -;
while(left <= right)
{
int mid = left + (right - left)/;
if(nums[mid] == target)
{
index = mid;
break;
}
else if( nums[mid] > target )
right = mid -;
else
left = mid + ;
}
// target doesn't exist in array
if( index == -)
{
ans.push_back(-);
ans.push_back(-);
return ans;
}
// target exist in array
int indexL = index;
int indexR = index;
while( nums[indexL] == target && indexL >= ) indexL--;
while( nums[indexR] == target && indexR < nums.size() ) indexR++;
int n = nums.size();
ans.push_back(indexL+);
ans.push_back(indexR-);
return ans;