leetcode 7. 在有序可重复数组旋转后搜索 Search in Rotated Sorted Array II

Search in Rotated Sorted Array II

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.

解题思路：
本题基于leetcode 6. 在有序数组旋转后搜索 Search in Rotated Sorted Array，允许有序数组中存在重复。
与上一题（不允许重复）的不同是，
nums[m] > nums[l] : (l, m-1)单增
nums[m] <= nums[l] : (m+1, r)不一定单增，因为{1,3,1,1,1} 或{1,1,1,3,1}

此时，可以将上述条件分开来看
nums[m] < nums[l] : (m+1, r)一定单增
num[m] == nums[l] : 将 l+1，重新递归计算 （当[l, r],将 r-1）

class Solution {
public:
    //nums 数组边界为 [l,r)
    bool searchR(vector<int>& nums,int l, int r, int target) {
        if (r <= l)
            return false;
        int m = (l+r) / 2;
        if (nums[m] == target)
            return true;

        if (nums[l] < nums[m]) {
            if (target >= nums[l] && target < nums[m])
                return searchR(nums, l, m, target);
            else
                return searchR(nums, m+1, r, target);
        } else if (nums[l] > nums[m]) {
            if(target > nums[m] && target <= nums[r-1])
                return searchR(nums, m+1, r, target);
            else
                return searchR(nums, l, m, target);    
        } else {
            return searchR(nums, ++l, r, target);    

        }
    }

    bool search(vector<int>& nums, int target) {
        return searchR(nums, 0, nums.size(), target);
    }
};