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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.
解法：同旋转数组查找I，本题也使用二分查找，只是在nums[mid]和nums[right]（或者nums[right]）比较时若二者相等则left++（或者right--）即可。nums[mid]与nums[left]比较的代码：
class Solution { public: int search(vector<int>& nums, int target) { int n = nums.size(); bool res = false; int left = 0, right = n - 1; while(left <= right) { int mid = ((left + right) >> 1); if(target == nums[mid]) { res = true; break; } else if(nums[mid] > nums[left]) //前半部分有序 { if(target >= nums[left] && target < nums[mid]) right = mid - 1; else left = mid + 1; } else if(nums[mid] < nums[left]) //后半部分有序 { if(target > nums[mid] && target <= nums[right]) left = mid + 1; else right = mid - 1; } else //nums[mid]==nums[left]的情况，包括mid=left和存在重复值两种情况 left++; } return res; } };
nums[mid]与nums[right]比较的代码：
class Solution { public: int search(vector<int>& nums, int target) { int n = nums.size(); bool res = false; int left = 0, right = n - 1; while(left <= right) { int mid = (left + right) >> 1; if(target == nums[mid]) { res = true; break; } else if(nums[mid] < nums[right]) //后半部分有序 { if(target > nums[mid] && target <= nums[right]) left = mid + 1; else right = mid - 1; } else if(nums[mid] > nums[right]) //前半部分有序 { if(target >= nums[left] && target < nums[mid]) right = mid - 1; else left = mid + 1; } else //存在重复值的情况 right--; } return res; } };
[LeetCode]26. Search in Rotated Array II旋转数组查找II
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原文地址：http://www.cnblogs.com/aprilcheny/p/4882635.html
