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leetcode 二分查找 Search in Rotated Sorted ArrayII

2024-07-26 12:15:37   222人阅读

Search in Rotated Sorted Array II

 Total Accepted: 18500 Total Submissions: 59945My Submissions

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.


题意：一个已经排序好的数组，被按某个位置旋转了一次，给定一个值target，在该旋转后的数组里查找该值。（数组中的元素可能重复）
思路：二分查找
难点在于确定往数组的哪一半段继续二分查找
设起点、中间点、终点分别为 start、middle、end (采用前闭后开的区间表示方法
如果target = A[middle] return middle
如果A[middle] > A[start]，则[start,middle)单调递增
1.如果target < A[middle] && target >= A[start]，则 end = middle
2.start = middle + 1, otherwise
如果A[middle] < A[start]，则[middle,end)单调递增
1.如果target > A[middle] && target <= A[end - 1]，则 start = middle + 1
2.end = middle, otherwise
如果A[middle] == A[start]，那A[start]也不会是target，可以通过start++; 去掉A[start]


复杂度：时间O(n)，空间O(1)


int search(int A[], int n, int target){
	int start = 0, end = n, middle ;
	while(start < end){
		middle = (start + end) / 2;
		if(A[middle] == target) return middle;
		if(A[middle] > A[start]){
			if(target >= A[start] && target < A[middle]){
				end = middle;
			}else{
				start = middle + 1;
			}
		}else if(A[middle] < A[start]){
			if(target > A[middle] && target <= A[end - 1]){
				start = middle + 1;
			}else{
				end = middle;
			}
		}
		else{
			++start;
		}
	}
	return -1;
}


leetcode 二分查找 Search in Rotated Sorted ArrayII


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