leetcode 二分查找 Search in Rotated Sorted Array

Search in Rotated Sorted Array

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Suppose a sorted array 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.

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。

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 (採用前闭后开的区间表示方法

假设target = A[middle] return middle

假设A[middle] >= A[start]，则[start,middle)单调递增

1.假设target < A[middle] && target >= A[start]，则 end = middle

2.start = middle + 1, otherwise

假设A[middle] < A[start]。则[middle,end)单调递增

1.假设target > A[middle] && target <= A[end - 1]，则 start = middle + 1

2.end = middle, otherwise





复杂度：时间O(log 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(target > A[middle] && target <= A[end - 1]){
				start = middle + 1;
			}else{
				end = middle;
			}
		}
	}
	return -1;
}