【LeetCode】C# 81、Search in Rotated Sorted Array II

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

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

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

The array may contain duplicates.

在翻转过的包含重复数的数组中查找元素。

思路一：直接遍历。也能AC……

public class Solution {    public bool Search(int[] nums, int target) {        for (int i = 0; i < nums.Length; i++)        {            if (nums[i] == target)                return true;        }        return false;    }}

思路二：
因为重复的原因，可能左右两端数值相同，使我们失去二分挑边的功能。解决办法是对边缘移动一步，直到边缘和中间不在相等或者相遇。所以最坏情况（比如全部都是一个元素，或者只有一个元素不同于其他元素，而他就在最后一个）就会出现每次移动一步，总共是n步，算法的时间复杂度变成O(n)。
思路参考：http://blog.csdn.net/linhuanmars/article/details/20588511

public bool search(int[] nums, int target) {    if(nums==null || nums.Length==0)        return false;    int l = 0;    int r = nums.Length-1;    while(l<=r)    {        int m = (l+r)/2;        if(nums[m]==target)            return true;        if(nums[m]>nums[l])        {            if(nums[m]>target && nums[l]<=target)            {                r = m-1;            }            else            {                l = m+1;            }        }        else if(nums[m]<nums[l])        {            if(nums[m]<target && nums[r]>=target)            {                l = m+1;            }            else            {                r = m-1;            }                        }        else        {            l++;        }    }    return false;}