33. Search in Rotated Sorted Array

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33. Search in Rotated Sorted Array

题目

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

解析

// 33. Search in Rotated Sorted Array
class Solution_33 {
public:
	// 本质:不管什么情况，都是只是low,high进行移动，二分查找时候一定记住要有常数步的前进，防止进入死循环

//input : [3, 1]
//       	1
//Output : -1 bug1:加等号
//	 Expected : 1
	int search(vector<int>& nums, int target) {
		if (nums.size()==1&&nums[0]==target)
		{
			return 0;
		}
		int low = 0, high = nums.size()-1;
		while (low<=high)
		{
			int mid = low + (high - low) / 2;
			if (nums[mid]==target)
			{
				return mid;
			}
			if (nums[mid]>target)
			{
				if (nums[low]<=nums[mid]) //低半部分有序;  bug 1有序部分要用等号
				{
					if (nums[low]<=target) //target在低半部分序列中
					{
						high = mid - 1;
					}
					else
					{
						low = mid + 1;
					}
				}
				else // 后半部分有序，且nums[mid]>target;必位于前半部分
				{
					high = mid - 1;
				}
			}
			else
			{
				if (nums[mid]<=nums[high]) //后半部分有序
				{
					if (nums[high]>=target)
					{
						low = mid + 1;
					}
					else
					{
						high = mid - 1;
					}
				}
				else //nums[mid]>target 且前部分有序
				{
					low = mid + 1;
				}
			}
			
		}
		return -1;
	}

	int search2(int A[], int n, int target) {

		int low = 0, high = n - 1;

		while (low<=high)
		{
			int mid = low + (high - low) >> 1;
			if (A[mid]==target)
			{
				return mid;
			}
			if (A[mid]>=A[low])  //低半部分有序；先比较区间，在比较关键字target
			{
				if (A[mid]>target&& target>=A[low]) //bug 2: 调整那个，就不用等号
				{
					high = mid - 1;
				}
				else
				{
					low = mid + 1;
				}
			}
			else //后半部分有序
			{
				if (A[mid]<target&&target<=A[high])
				{
					low = mid + 1;
				}
				else
				{
					high = mid - 1;
				}
			}
		}
		return -1;
	}

	int search_ref(vector<int>& nums, int target) {
		int l = 0, r = nums.size() - 1;
		while (l <= r) {
			int mid = (l + r) / 2;
			if (target == nums[mid])
				return mid;
			// there exists rotation; the middle element is in the left part of the array
			if (nums[mid] > nums[r]) {
				if (target < nums[mid] && target >= nums[l])
					r = mid - 1;
				else
					l = mid + 1;
			}
			// there exists rotation; the middle element is in the right part of the array
			else if (nums[mid] < nums[l]) {
				if (target > nums[mid] && target <= nums[r])
					l = mid + 1;
				else
					r = mid - 1;
			}
			// there is no rotation; just like normal binary search
			else {
				if (target < nums[mid])
					r = mid - 1;
				else
					l = mid + 1;
			}
		}
		return -1;
	}

};

题目来源

• 33. Search in Rotated Sorted Array