[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索 II

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.

33. Search in Rotated Sorted Array 的拓展，数组中允许出现重复数字，这个也会影响我们选择哪半边继续搜索，之前判断左右部分是否有序的方法就失效了，因为可能有这种58555情况，虽然起点小于等于中间，但不代表右边就不是有序的，因为中点也小于等于终点，所有右边也是有序的。所以，如果遇到这种中点和两边相同的情况，我们两边都要搜索。

Java：

public class Solution {
    public boolean search(int[] nums, int target) {
        return helper(nums, 0, nums.length - 1, target);
    }
    
    public boolean helper(int[] nums, int min, int max, int target){
        int mid = min + (max - min) / 2;
        // 不符合min <= max则返回假
        if(min > max){
            return false;
        }
        if(nums[mid] == target){
            return true;
        }
        boolean left = false, right = false;
        // 如果左边是有序的
        if(nums[min] <= nums[mid]){
            // 看目标是否在左边有序数列中
            if(nums[min] <= target && target < nums[mid]){
                left = helper(nums, min, mid - 1, target);
            } else {
                left = helper(nums, mid + 1, max, target);
            }
        }
        // 如果右边也是有序的
        if(nums[mid] <= nums[max]){
            // 看目标是否在右边有序数列中
            if(nums[mid] < target && target <= nums[max]){
                right = helper(nums, mid + 1, max, target);
            } else {
                right = helper(nums, min, mid - 1, target);
            }
        }
        // 左右两边有一个包含目标就返回真
        return left || right;
    }
}

Python：

class Solution(object):
    def search(self, nums, target):
        left, right = 0, len(nums) - 1
        
        while left <= right:
            mid = left + (right - left) / 2
            
            if nums[mid] == target:
                return True
            elif nums[mid] == nums[left]:
                left += 1
            elif (nums[mid] > nums[left] and nums[left] <= target < nums[mid]) or 
                 (nums[mid] < nums[left] and not (nums[mid] < target <= nums[right])):
                right = mid - 1
            else:
                left = mid + 1

        return False

　　

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[LeetCode] 33. Search in Rotated Sorted Array 在旋转有序数组中搜索

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