Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

原创最新推荐文章于 2022-05-26 16:47:57 发布 · 237 阅读

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本文探讨了在包含重复值的旋转排序数组中查找最小元素的算法，详细介绍了如何在存在重复元素的情况下调整二分查找策略，并分析了运行时复杂性的变化。

Problem

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

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

Suppose a sorted array 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).

Find the minimum element.

The array may contain duplicates.

Solution

就多了一个当 nums[mid] == nums[right] 的一个判断。

只能一步一步向右走来判断是否一样。。。。

class Solution {
public:
    int findMin(vector<int>& nums) {
       int left = 0, right = nums.size() -1;
        
        while( left < right) {
            int mid = (left + right) /2;
            if(nums[mid] < nums[right]){
                right = mid;
            }
            else if(nums[mid] > nums[right]) {
                left = mid + 1;
            }
            else {
                bool rightSame = true;
                for( int i = mid + 1; i < right; i++){
                    if( nums[i] != nums[i-1]){
                        rightSame = false; break;
                    }
                }
                rightSame ? right = mid : left = mid + 1; 
                
            }
        }
        return nums[left];
                
    }
};