C++实现LeetCode(81.在旋转有序数组中搜索之二) C++实现LeetCode(81.在旋转有序数

[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索之二 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:

【C++实现LeetCode(81.在旋转有序数组中搜索之二)】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& 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; }};

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