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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.
这题跟上一题一样，都是二分查找法，但是需要思考，对于时间复杂度，存在重复元素会有影响吗？可以考虑如[1,0,1,1,...,1]这样的数组，时间复杂度最坏会 变成O（N）。
class Solution { public: int search(vector<int>& nums, int target) { int left = 0, right = nums.size()-1; while(left <= right) { int middle = left + ((right-left)>>1); if (nums[middle] == target) return true; if (nums[middle] < nums[right]) { if (nums[middle] < target && target <= nums[right])left = middle + 1; else right = middle - 1; } else if (nums[middle] > nums[right]) { if (nums[left] <= target && target < nums[middle])right = middle - 1; else left = middle + 1; } else right--; } return false; } };
【LeetCode】81. Search in Rotated Sorted Array II
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原文地址：http://www.cnblogs.com/Doctengineer/p/5813035.html
