81. Search in Rotated Sorted Array II

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mb5fca0be3cc41d 2019-09-12 11:30:00

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?

class Solution {
    public boolean search(int[] nums, int target) {
        int n = nums.length;
        int l = 0, r = n - 1;
        while(l <= r){
            int mid = l + (r-l)/2;
            if(nums[mid] == target) return true;
            if(nums[mid] < nums[r]){//大条件，说明mid到r是升序
                if(target > nums[mid] && target <= nums[r]) l = mid+1;//需要同时满足这两个条件
                else r = mid-1;
            }
            else if(nums[mid] > nums[r]){
                if(target < nums[mid] && target >= nums[l]) r = mid - 1;
                else l = mid + 1;
            }
            else r--;//中间值等于最右值时，目标值既可以在左边又可以在右边，既然我们比较的是r，那就r--
        }
        return false;
    }
}