LeetCode 81. Search in Rotated Sorted Array II

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本文介绍了一个在含有重复元素的旋转有序数组中查找特定目标值的算法。通过调整二分查找法来解决因数组旋转和重复元素导致的问题，并讨论了其最坏情况下的时间复杂度为O(N)。

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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.

Worst Time Complexity would be O(N).... for example: whole array is [4, 4, 4, 4, 4.....4] then the target is 3.

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


bool search(vector<int>& nums, int target) {
    if(nums.size() == 0) return false;
    int low = 0;
    int high = nums.size() - 1;
    while(low <= high) {
        int mid = low + (high - low) / 2;
        if(nums[mid] == target) return true;
        if(nums[mid] < nums[high]) {
            if(target <= nums[high] && target > nums[mid]) {
                low = mid + 1;
            } else high = mid - 1;

        } else if(nums[mid] > nums[high]) {
            if(target >= nums[low] && target < nums[mid]) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        } else { high--;}
    }
    return false;
}

// duplicates in rotated sorted array.
// For example: 4, 4, 5, 6, 1, 2, 3, 4, target is 3.
int main(void) {
    vector<int> nums{4, 4, 5, 6, 1, 2, 3, 4};
    int target = 2;
    bool found = search(nums, target);
    cout << found << endl;
}