LeetCode Search for a Range (二分查找)

题意

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].

给定一个有序数组和一个目标值，查找出目标值在数组中出现的发生下标区域。时间复杂度要求log(N)。

解法

刚开始以为下标也需要用二分来查找，但看讨论发现大家都是找出目标值然后再向两边搜索-_-b，这样做的话，最差的情况下（数组里的数全都一样）就是O(N)了，但是也能过。

class Solution
{
public:
    vector<int> searchRange(vector<int>& nums, int target)
    {
	    int	left = 0;
	    int	right = nums.size() - 1;
	    int	index_a = -1;
	    int	index_b = -1;
	    int	index = 0;
	    bool	flag = false;
	    while(left <= right)
	    {
		    int	mid = (left + right) >> 1;
		    if(nums[mid] == target)
		    {
			    flag = true;
			    index = mid;
			    break;
		    }
		    else
			    if(nums[mid] < target)
				    left = mid + 1;
			    else
				    right = mid - 1;
	    }
	    if(flag)
		    index_a = index;
	    while(index_a >= 1 && nums[index_a - 1] == target)
		    index_a --;
	    if(flag)
		    index_b = index;
	    while(index_b < nums.size() - 1 && nums[index_b + 1] == target)
		    index_b ++;
	    vector<int>	rt = {index_a,index_b};
	    return	rt;
    }
};