leetcode 34. Search for a Range

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本文介绍了一种O(log n)复杂度的算法，用于在已排序的整数数组中找到指定目标值的起始和结束索引。如果目标值不存在，则返回[-1, -1]。

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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].

int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
	if (numsSize == 0 || nums == NULL)
	{
		*returnSize = 2;
		int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
		re[0] = -1; re[1] = -1;
		return re;
	}
	if (nums[0] > target || nums[numsSize - 1] < target)
	{
		*returnSize = 2;
		int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
		re[0] = -1; re[1] = -1;
		return re;
	}
	if (numsSize == 1)
	{
		*returnSize = 2;
		int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
		if (nums[0] == target)
		{
			re[0] = 0; re[1] = 0; return re;
		}
		else
		{
			re[0] = -1; re[1] = -1;
			return re;
		}
	}
	int st1 = 0; int end1 = numsSize - 1;
	while (end1 > st1)
	{
		if (nums[end1] > target)
			end1 = (st1 + end1) / 2;
		else
		{
			end1 = 2 * end1 - st1 > numsSize - 1 ? numsSize - 1 : 2 * end1 - st1;
			st1 = (st1 + end1) / 2;
		}
		if (nums[end1] == target&&end1 == numsSize - 1)
			break;
	}
	if (nums[end1] != target)
	{
		*returnSize = 2;
		int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
		re[0] = -1; re[1] = -1;
		return re;
	}
	int st2 = 0; int end2 = end1;
	if (nums[0] == target)
		end2 = 0;
	else
		while (true)
		{
			if (nums[end2] == target&&nums[st2] != target && (end2 - st2 == 1 || end2 == st2))
				break;
			if (nums[end2] >= target)
				end2 = (st2 + end2) / 2;
			else
			{
				end2 = 2 * end2 - st2 > end1 ? end1 : 2 * end2 - st2;
				st2 = (st2 + end2) / 2;
			}
		}
	*returnSize = 2;
	int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
	re[0] = end2; re[1] = end1;
	return re;
}

accepted