LeetCode-34. Search for a Range

LeetCode-34. Search for a Range

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

解题思路：可以用两次二分查找分别找出上限和下限

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

也可以利用STL自带的函数

vector<int> searchRange(vector<int>& nums, int target) {
    auto bounds = equal_range(nums.begin(), nums.end(), target);
    if (bounds.first == bounds.second)
        return {-1, -1};
    return {bounds.first - nums.begin(), bounds.second - nums.begin() - 1};
}

equal_range是C++ STL中的一种二分查找的算法，试图在已排序的[first,last)中寻找value，它返回一对迭代器i和j，其中i是在不破坏次序的前提下，value可插入的第一个位置（亦即lower_bound），j则是在不破坏次序的前提下，value可插入的最后一个位置（亦即upper_bound），因此，[i,j)内的每个元素都等同于value，而且[i,j)是[first,last)之中符合此一性质的最大子区间。

vector<int> searchRange(vector<int>& nums, int target) {
    int lo = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
    if (lo == nums.size() || nums[lo] != target)
        return {-1, -1};
    int hi = upper_bound(nums.begin(), nums.end(), target) - nums.begin() - 1;
    return {lo, hi};
}