【LeetCode】4.Search in Rotated Sorted Array II

题目描述(Medium)

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
You are given a target value to search. If found in the array return true, otherwise return false.

题目链接

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/description/

Example 1:

 Input: nums = [2,5,6,0,0,1,2], target = 0
 Output: true

Example 2:

 Input: nums = [2,5,6,0,0,1,2], target = 3
 Output: false

算法分析

二分查找，需要判断好左右边界，这里主要是先判断出递增子序列，然后以递增子序列作为判断条件，确定左右边界；

若出现A[mid] == A[first]时，不能确定递增的情况，则++first跳过该重复元素。

提交代码：

class Solution {
public:
	bool search(vector<int>& nums, int target) {
		if (nums.empty()) return false;

		int beg = 0, end = nums.size() - 1;
		int mid;

		while (beg <= end)
		{
			mid = (beg + end) / 2;
			if (nums[mid] == target)
				return true;

			if (nums[mid] > nums[beg])
			{
				if (target >= nums[beg] && target < nums[mid])
					end = mid - 1;
				else
					beg = mid + 1;
			}
			else if (nums[mid] < nums[beg])
			{
				if (target > nums[mid] && target <= nums[end])
					beg = mid + 1;
				else
					end = mid - 1;
			}
			else
				//跳过重复元素
				++beg;
		}
		return false;
	}
};

测试代码：

// ====================测试代码====================
void Test(const char* testName, vector<int>& nums, int target, bool expected)
{
	if (testName != nullptr)
		printf("%s begins: \n", testName);

	Solution s;
	bool result = s.search(nums, target);

	if (result == expected)
		printf("passed\n");
	else
		printf("failed\n");

}

int main(int argc, char* argv[])
{

	vector<int> array1 = { 2, 5, 6, 0, 0, 1, 2 };
	Test("Test1", array1, 0, true);

	vector<int> array2 = { 2, 5, 6, 0, 0, 1, 2 };
	Test("Test2", array2, 3, false);

	vector<int> array3 = { 2, 5, 6, 0, 0, 1, 2 };
	Test("Test3", array3, 2, true);

	vector<int> array4 = { 1, 3, 1, 1, 1 };
	Test("Test4", array4, 3, true);

	return 0;
}