題目

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

解題之法

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

分析

這道是之前那道 Search in Rotated Sorted Array 在旋轉(zhuǎn)有序數(shù)組中搜索 的延伸，現(xiàn)在數(shù)組中允許出現(xiàn)重復(fù)數(shù)字，這個(gè)也會(huì)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?cè)诒容^中間值和最右值時(shí)就完全符合之前所說(shuō)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(fù)值，就會(huì)出現(xiàn)來(lái)面兩種情況，[3 1 1] 和 [1 1 3 1]，對(duì)于這兩種情況中間值等于最右值時(shí)，目標(biāo)值3既可以在左邊又可以在右邊，那怎么辦么，對(duì)于這種情況其實(shí)處理非常簡(jiǎn)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 在旋轉(zhuǎn)有序數(shù)組中搜索 中的方法。