Search in Rotated Sorted Array II

描述

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.

分析

允许重复元素，则上一题中如果A[left] <= A[mid],那么[left,mid]为递增序列的假设就不能成立了，比如[1,3,1,1,1]。

既然A[left] <= A[mid]不能确定递增，那就把它拆分成两个条件：

• 若A[left] < A[mid]，则区间[left,mid]一定递增
• 若A[left] == A[mid] 确定不了，那就left++，往下看一步即可。

代码

Python

// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
public class Solution {
    public boolean search(int[] nums, int target) {
        int first = 0, last = nums.length;
        while (first != last) {
            final int mid = first  + (last - first) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[first] < nums[mid]) {
                if (nums[first] <= target && target < nums[mid])
                    last = mid;
                else
                    first = mid + 1;
            } else if (nums[first] > nums[mid]) {
                if (nums[mid] < target && target <= nums[last-1])
                    first = mid + 1;
                else
                    last = mid;
            } else
                //skip duplicate one
                first++;
        }
        return false;
    }
};

Java

// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
class Solution {
public:
    bool search(const vector<int>& nums, int target) {
        int first = 0, last = nums.size();
        while (first != last) {
            const int mid = first  + (last - first) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[first] < nums[mid]) {
                if (nums[first] <= target && target < nums[mid])
                    last = mid;
                else
                    first = mid + 1;
            } else if (nums[first] > nums[mid]) {
                if (nums[mid] < target && target <= nums[last-1])
                    first = mid + 1;
                else
                    last = mid;
            } else
                //skip duplicate one
                first++;
        }
        return false;
    }
};

C++

# Search in Rotated Sorted Array II
# Time Complexity: O(n), Space Complexity: O(1)
class Solution:
    def search(self, nums: list[int], target: int) -> bool:
        first = 0
        last = len(nums)
        while first != last:
            mid = first + (last - first) // 2
            if nums[mid] == target:
                return True
            if nums[first] < nums[mid]:
                if nums[first] <= target and target < nums[mid]:
                    last = mid
                else:
                    first = mid + 1
            elif nums[first] > nums[mid]:
                if nums[mid] < target and target <= nums[last-1]:
                    first = mid + 1
                else:
                    last = mid
            else:
                #skip duplicate one
                first += 1
        return False

相关题目

• Search in Rotated Sorted Array
• Find Minimum in Rotated Sorted Array
• Find Minimum in Rotated Sorted Array II