leetcode 81: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.

思路：

思路和leetcode 33:Revised Binary Search类似，只不过需要考虑重复的情况。

比如：

11111311111111111

如果照以前的方法，low,mid,high都是1，不知那边是排好序的哪边是没有排好序的。

但是这样情况下，[low,mid]与[mid,high]两者之间必有一个是全为1的，因此需要一个judge函数来判断哪边全为1.

实现如下：

class Solution {public:bool search(vector<int>& nums, int target) {int size = nums.size();if (size == 0) return 0;int low = 0, high = nums.size() - 1;return notBinary(nums, low, high, target);}bool binary(vector<int>&nums, int low, int high, int target){if (low == high && nums[low] == target) return 1;else if (low == high) return 0;int mid = (low + high) / 2;if (nums[mid] == target) return 1;else if (nums[mid] > target&&mid>low) return binary(nums, low, mid - 1, target);else if (mid<high) return binary(nums, mid + 1, high, target);}bool notBinary(vector<int>&nums, int low, int high, int target){if (low == (high - 1)) return nums[low] == target ? 1 : (nums[high] == target ? 1 : 0);if (low == high && nums[low] == target) return 1;else if (low == high) return 0;int mid = (low + high) / 2;if (nums[mid] == target) return 1;if (target >= nums[low] && target < nums[mid] && mid>low) return binary(nums, low, mid - 1, target);else if (target>nums[mid] && target <= nums[high] && mid<high)return binary(nums, mid + 1, high, target);else if (nums[low] == nums[mid])                //新增的{if (judge(nums,low,mid) && mid<high) return notBinary(nums, mid + 1, high, target);else if (judge(nums,mid,high) && mid>low) return notBinary(nums, low, mid - 1, target);else return 0;}else if (nums[low] < nums[mid] && mid<high)return notBinary(nums, mid + 1, high, target);else if (mid>low) return notBinary(nums, low, mid - 1, target);else return 0;}bool judge(vector<int>&nums, int low, int high){if (low == high) return true;while (low < high){if (nums[low] != nums[low + 1]) return false;low++;}return true;}};

下面提供一种更为简洁的方法，思路和上面类似，只不过对nums[mid]=nums[high]的处理更为巧妙。

时间复杂度：O（lgn）

class Solution {public:bool search(vector<int> &A, int target) {int n = A.size();int lo = 0, hi = n - 1;int mid = 0;while (lo<hi){mid = (lo + hi) / 2;if (A[mid] == target) return true;if (A[mid]>A[hi]){if (A[mid]>target && A[lo] <= target) hi = mid;else lo = mid + 1;}else if (A[mid] < A[hi]){if (A[mid]<target && A[hi] >= target) lo = mid + 1;else hi = mid;}else hi--;}return A[lo] == target ? true : false;}};