[LeetCode] 81. Search in Rotated Sorted Array II

[grandyang]

 

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?

 

这道是之前那道 Search in Rotated Sorted Array 的延伸，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，可以得到代码如下：

 

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int n = nums.size(), left = 0, right = n - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) return true;
            if (nums[mid] < nums[right]) {
                if (nums[mid] < target && nums[right] >= target) left = mid + 1;
                else right = mid - 1;
            } else if (nums[mid] > nums[right]){
                if (nums[left] <= target && nums[mid] > target) right = mid - 1;
                else left = mid + 1;
            } else --right;
        }
        return false;
    }
};

 

Github 同步地址：

#81

 

类似题目：

Search in Rotated Sorted Array

 

参考资料：

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

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/discuss/28194/C%2B%2B-concise-log(n)-solution

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/discuss/28218/My-8ms-C%2B%2B-solution-(o(logn)-on-average-o(n)-worst-case)

 

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