LeetCode——Search in Rotated Sorted Array

标签：leetcode

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.

题目：假设一个有序数组在某一点上被折叠，你不知道哪一点上。

给定一个数组和一个目标值，找出目标值在数组中的索引。

思路：可以用线性查找的方法，但是就背离了题目中的有序这个条件。比较好的办法是二分查找，但是需要找到有序的一部分，才可以使用二分。

	public int search(int[] A, int target) {
		int low = 0, high = A.length - 1;
		while (low <= high) {
			int mid = (high + low) / 2;
			if (target == A[mid])
				return mid;
			if (A[mid] > A[low]) {
				if (target >= A[low] && target < A[mid])
					high = mid - 1;
				else
					low = mid + 1;
			} else if(A[mid] < A[low]){
				if (target <= A[high] && target > A[mid])
					low = mid + 1;
				else
					high = mid - 1;
			}else
				low += 1;
		}
		return -1;
	}

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LeetCode——Search in Rotated Sorted Array

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/38141433