[LeetCode] 81. Search in Rotated Sorted Array II 在旋轉(zhuǎn)有序數(shù)組中搜索之二

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:

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?

這道是之前那道 Search in Rotated Sorted Array 的延伸，現(xiàn)在數(shù)組中允許出現(xiàn)重復數(shù)字，這個也會影響我們選擇哪半邊繼續(xù)搜索，由于之前那道題不存在相同值，我們在比較中間值和最右值時就完全符合之前所說的規(guī)律：如果中間的數(shù)小于最右邊的數(shù)，則右半段是有序的，若中間數(shù)大于最右邊數(shù)，則左半段是有序的。而如果可以有重復值，就會出現(xiàn)來面兩種情況，[3 1 1] 和 [1 1 3 1]，對于這兩種情況中間值等于最右值時，目標值3既可以在左邊又可以在右邊，那怎么辦么，對于這種情況其實處理非常簡單，只要把最右值向左一位即可繼續(xù)循環(huán)，如果還相同則繼續(xù)移，直到移到不同值為止，然后其他部分還采用 Search in Rotated Sorted Array 中的方法，可以得到代碼如下：

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;
    }
};

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