81. Search in Rotated Sorted Array II

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?

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.

解析

  • 和Search in Rotated Sorted Array唯一的区别是这道题目中元素会有重复的情况出现。不过正是因为这个条件的出现,出现了比较复杂的case,甚至影响到了算法的时间复杂度。原来我们是依靠中间和边缘元素的大小关系,来判断哪一半是不受rotate影响,仍然有序的。而现在因为重复的出现,如果我们遇到中间和边缘相等的情况,我们就丢失了哪边有序的信息,因为哪边都有可能是有序的结果。假设原数组是{1,2,3,3,3,3,3},那么旋转之后有可能是{3,3,3,3,3,1,2},或者{3,1,2,3,3,3,3},这样的我们判断左边缘和中心的时候都是3,如果我们要寻找1或者2,我们并不知道应该跳向哪一半。解决的办法只能是对边缘移动一步,直到边缘和中间不在相等或者相遇,这就导致了会有不能切去一半的可能。

 // The array may contain duplicates.
    bool search(vector<int>& nums, int target) {
if (nums.empty())
        {
            return false;
        }
        int low = 0, high = nums.size() - 1;
        int mid = 0;
        while (low<=high)
        {
            mid = low + (high - low) / 2;
            if (nums[mid]==target)
            {
                return true;    
            }
            if (nums[mid]>nums[high]) // 前半部分有序;后半部分无序
            {
                if (nums[mid]>target&&nums[low]<=target)
                {
                    high = mid-1;
                }
                else
                {
                    low = mid + 1;
                }
            }
            else if (nums[mid]<nums[high]) // 后半部分有序
            {
                if (nums[mid]<target&&target<=nums[high])
                {
                    low = mid+1;
                }
                else
                {
                    high = mid-1;
                }
            }
            else
            {
                high--; //与high在比较
            }
        }
        return false;
    }

// 81. Search in Rotated Sorted Array II
class Solution_81 {
public:
    // The array may contain duplicates.
    bool search_(vector<int>& nums, int target) {

        if (nums.empty())
        {
            return false;
        }
        int low = 0, high = nums.size() - 1;
        int mid = 0;
        while (low<high)
        {
            mid = low + (high - low) / 2;
            if (nums[mid]==target)
            {
                return true;    
            }
            if (nums[mid]>nums[high]) // 前半部分有序;后半部分无序
            {
                if (nums[mid]>target&&nums[low]<=target)
                {
                    high = mid;
                }
                else
                {
                    low = mid + 1;
                }
            }
            else if (nums[mid]<nums[high]) // 后半部分有序
            {
                if (nums[mid]<target&&target<=nums[high])
                {
                    low = mid+1;
                }
                else
                {
                    high = mid;
                }
            }
            else
            {
                high--;
            }
        }
        return nums[low] == target ? true : false;
    }

    bool search(int A[], int n, int target) {
        vector<int> vec(A, A + n);
        return search_(vec, target);
    }
};

题目来源

    原文作者:ranjiewen
    原文地址: https://www.cnblogs.com/ranjiewen/p/8734888.html
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