81.Search in Rotated Sorted Array II

Tags: medium Array

Link: https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

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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.

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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?

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Answer:

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int middle = left + ((right - left) >> 1);
            if (nums[middle] == target) return true;
            else if (nums[middle] < nums[right]) { //[middle, right] 有序
                if (nums[middle] < target && target <= nums[right]) left = middle + 1;
                else right = middle - 1;
            }
            else if (nums[middle] > nums[right]) { //[left, middle]有序
                if (nums[left] <= target && target < nums[middle]) right = middle - 1;
                else left = middle + 1;
            }
            else --right;
        }

        return false;
    }
};

现在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，由于之前那道题不存在相同值，我们在比较中间值和最右值时就完全符合之前所说的规律： 如果中间的数小于最右边的数，则右半段是有序的，若中间数大于最右边数，则左半段是有序的 。而如果可以有重复值，就会出现来面两种情况，[3 1 1] 和 [1 1 3 1]，对于这两种情况中间值等于最右值时，目标值3既可以在左边又可以在右边，那怎么办么，对于这种情况其实处理非常简单，只要把最右值向左一位即可继续循环，如果还相同则继续移，直到移到不同值为止，然后其他部分还采用 Search in Rotated Sorted Array 中的方法，