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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.
分析：当允许有重复元素出现的时候，那么Search in Rotated Sorted Array中line14：nums[low] <= nums[mid]所能决定的序列就并非单调递增序列，例如[3,1,3,3,3]。在这种情形下，只需将原来的条件拆分为两种情形：
1. nums[low] < nums[mid] 那么之间的序列一定是递增的，代码为Search in Rotated Sorted Array中的line14~17
2. nums[low] == nums[mid] 由于不能够判断数字到底在何处增加、减少，进行low++后观察。因为low++之后nums[low]一定会变大或者保持不变，变大即满足nums[low] > nums[mid]，不变则表示已经进入到重复数字区域。
运行时间 11ms
1 class Solution { 2 public: 3 bool search(vector<int>& nums, int target) { 4 if(nums.size() == 0) return false; 5 if(nums.size() == 1){ 6 if(nums[0] == target) return true; 7 else return false; 8 } 9 10 int low = 0, high = nums.size()-1; 11 while(low <= high){ 12 int mid = (low + high) / 2; 13 if(nums[mid] == target) return true; 14 if(nums[low] < nums[mid]){ 15 if(nums[low] <= target && target < nums[mid]) high = mid - 1; 16 else low = mid + 1; 17 } 18 else if(nums[low] > nums[mid]){ 19 if(nums[mid] < target && target <= nums[high]) low = mid + 1; 20 else high = mid - 1; 21 } 22 else low++; 23 } 24 return false; 25 } 26 };
Search in Rotated Sorted Array II
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原文地址：http://www.cnblogs.com/amazingzoe/p/4466444.html
