Leetcode 81. Search in Rotated Sorted Array II

原题：

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.

解决方法：
由于重复数据的存在，这道题的逻辑更加复杂一些，通过mid 和hi的比较，可以确定mid是在哪一个区间里面：

- 如果mid 大于hi，说明mid是在原来的递增区间，然后比较target和low与mid的关系，确定是移动lo还是hi;

- 如果mid 小于hi, 说明mid是在旋转过去的递增区间，然后比较target和hi与mid的关系，确定是移动lo还是hi;

- 如果mid等于hi,我们就只能将hi往前移动1位。

代码：

bool search(vector<int>& nums, int target) {
if (nums.empty())
return false;

int lo = 0, hi = nums.size() -1;
while(lo <= hi){
int mid = lo + ( (hi -lo) >> 1 );
if (nums[mid] == target)
return true;

if (nums[mid] > nums[hi]){
if (nums[mid] > target && nums[lo] <= target){
hi = mid - 1;
}else{
lo = mid + 1;
}
}else if (nums[mid] < nums[hi]){
if (target > nums[mid] && target <= nums[hi]){
lo = mid + 1;
}else{
hi = mid - 1;
}
}else{
--hi;
}
}

return nums[lo] == target;
}