Description

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.

Solution

Binary Search, time O(logn) worst O(n), space O(1)

有点tricky的题，因为可能出现nums[left] == nums[mid] == nums[right]的情况，这种情况下只能遍历查找。

而且二分的条件也要改，不是根据nums[mid]和target的比较来决定，而是根据nums[left]和nums[mid]的关系来决定！

class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left <= right) {
            int mid = (left + right) / 2;
            
            if (target == nums[mid]) {
                return true;
            } else if (nums[left] < nums[mid]) {
                // nums[left, mid] is sorted
                if (target >= nums[left] && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else if (nums[left] > nums[mid]) {
                // nums[mid, right] is sorted
                if (target > nums[mid] && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                ++left;
            }
        }
        
        return false;
    }
}

也可以比较nums[mid]和nums[right]：

class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length < 1) {
            return false;
        }
        
        int left = 0;
        int right = nums.length - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return true;
            } else 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 {
                --right;
            }
        }
        
        return false;
    }
}