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leetcode 154. Find Minimum in Rotated Sorted Array II --------- java

2024-08-21 23:17:50   220人阅读

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.

 

接着153题，在数组中有重复的数字。

只需要判断一下当nums[left] == nums[mid] 即可。

public class Solution {
    public int findMin(int[] nums) {
        
        if( nums.length == 1)
            return nums[0];
        int left = 0,right = nums.length-1;
        int mid = (left+right)/2;
        int first = nums[0];

        while( left < right ){


            if( left == right-1 ){
                left++; 
            }
            else if( nums[mid] == nums[left]){
                left++;
            }else if( nums[left] < nums[mid] ){
                left = mid;
            }else{
                right = mid;
            }
            mid = (left+right)/2;
            first = Math.min(first,nums[left]);

        }
        return Math.min(first,nums[left]);
    }
}

 

leetcode 154. Find Minimum in Rotated Sorted Array II --------- java


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