154. Find Minimum in Rotated Sorted Array II

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于 2016-10-13 09:36:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Machelsky/p/5955243.html

探讨了在允许重复元素的旋转排序数组中查找最小值的方法。通过夹逼策略优化搜索过程，降低时间复杂度。

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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.

思路：duplicate采取夹逼。其他跟I一样。3 3 1 3 -》夹逼3 1 3 -> 3 1 -> 1

public class Solution {
    public int findMin(int[] nums) {
    if(nums==null||nums.length==0)return 0;
    int low=0,high=nums.length-1;
    while(low<high&&nums[low]>=nums[high]){
        int mid=low+(high-low)/2;
        if(nums[mid]<nums[high]){
            high=mid;
        }else if(nums[mid]>nums[high]){
            low=mid+1;
        }else{
            if(nums[mid]==nums[low])low++;
            else if(nums[mid]==nums[high])high--;
        }
    }
    return nums[low];
}
}