leetcode 81. Search in Rotated Sorted Array II

/* leetcode 81. Search in Rotated Sorted Array II Search in Rotated Sorted Array : http://blog.csdn.net/charles1e/article/details/51757259 题目大意：在上题的基础上，允许重复元素 解题思路：参考上一篇的，在这个基础上，如果nums[mid]==nums[right],就--right,忽略这个元素。 */

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    bool search(vector<int>& nums, int target) 
    {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right)
        {
            int mid = (left + right)/2;
            if (nums[mid] == target)
                //return mid;
                return true;
            if (nums[mid] < nums[right])
            {
                if (target > nums[mid] && target <= nums[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else if (nums[mid] > nums[right])
            {
                if (target >= nums[left] && target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else
            {
                //这是nums[mid]==nums[right]的情况，直接忽略即可
                right--;
            }
        }

        return false;
    }
};

void test_search()
{
    Solution sol;
    vector<int> nums{ 5,6,6,7,7,1,1,1,2,2,3,4 };
    for (int i = 1; i < 10; ++i)
        cout << i << " : " << sol.search(nums, i) << endl;
}

int main()
{
    test_search();

    return 0;
}