[LeetCode]Find Minimum in Rotated Sorted Array II

标签：java   leetcode   二分法   

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.

public class Solution {
	public int findMin(int[] num) {
		if(num.length==0){
			return Integer.MAX_VALUE;
		}
		int left = 0,right = num.length-1;
		while(left<right){
			if(num[left]<num[right]) return num[left];
			int mid = (left+right)/2;
			if(num[mid]>num[right]){
				if(mid==left) left = mid++;
				left = mid;
			}else if(num[mid]==num[right]){
				if(num[mid]==num[left]){
					return Math.min(findMin(Arrays.copyOfRange(num, left, mid)), findMin(Arrays.copyOfRange(num, mid+1,right+1)));
				}else if(num[mid]>num[left]){
					return num[left];
				}else {
					right = mid;
				}
					
			}else{
				right = mid;
			}
		}
		return num[left];
	}
}

public class Solution {
	public int findMin(int[] num) {
		assert num.length>0;
		int left = 0,right = num.length-1;
		while(left<right&&num[left]>=num[right]){
			int mid = (left+right)/2;
			if(num[mid]>num[right]){
				left = mid+1;
			}else if(num[mid]<num[right]){
				right = mid;
			}else{
				//
				left++;
			}
		}
		return num[left];
	}
}

[LeetCode]Find Minimum in Rotated Sorted Array II

标签：java   leetcode   二分法   

原文地址：http://blog.csdn.net/guorudi/article/details/42425343