LeetCode | Search in Rotated Sorted Array II

题目

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.

Search in Rotated Sorted Array里那套代码不好使了，因为重复元素造成无法判断究竟该向左还是向右，这时候只能用O(n)的方式慢慢找。

代码

public class SearchInRotatedSortedArrayII {
	public boolean search(int[] A, int target) {
		if (A == null || A.length == 0) {
			return false;
		}
		int low = 0;
		int high = A.length - 1;
		while (low <= high) {
			int mid = low + (high - low) / 2;
			if (target == A[mid]) {
				return true;
			}
			if (A[low] < A[mid]) {// left half is sorted
				if (A[low] <= target && target < A[mid]) {
					high = mid - 1;
				} else {
					low = mid + 1;
				}
			} else if (A[low] > A[mid]) { // right half is sorted
				if (A[mid] < target && target <= A[high]) {
					low = mid + 1;
				} else {
					high = mid - 1;
				}
			} else {
				++low;
			}
		}
		return false;
	}
}

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LeetCode | Search in Rotated Sorted Array II

原文：http://blog.csdn.net/perfect8886/article/details/22819531