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].
分析：题意为在一个有序数组中找到给定目标值的起始位置并返回，如果目标值不存在则返回[1,1].
思路：使用binarySearchLow()去找到不小于目标值数字的最小索引，使用binarySearchUp()去找到不大于目标值数字的最大索引，然后即可得到索引范围。
code如下：
class Solution {
private:
int binarySearchLow(vector<int>& nums, int target, int begin, int end)
{
if(begin > end) return begin;
int mid = begin + (end - begin) / 2;
if(nums[mid] < target) return binarySearchLow(nums, target, mid + 1, end);
else return binarySearchLow(nums, target, begin, mid - 1);
}
int binarySearchUp(vector<int>& nums, int target, int begin, int end)
{
if(begin > end) return end;
int mid = begin + (end - begin) / 2;
if(nums[mid] > target) return binarySearchUp(nums, target, begin, mid - 1);
else return binarySearchUp(nums, target, mid + 1, end);
}
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res(2, -1);
if(nums.empty()) return res;
int high = binarySearchUp(nums, target, 0, nums.size() -1);
int low = binarySearchLow(nums, target, 0, nums.size() - 1);
if(high >= low)
{
res[0] = low;
res[1] = high;
return res;
}
return res;
}
};
leetcode:Search for a Range（数组，二分查找）
原文：http://www.cnblogs.com/carsonzhu/p/4857327.html