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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.
Analysis:
For this problem, if we find that num[start]==num[mid]==num[end], we have to search both the left part and right part, because we do not know the min value is in which part. For example: 4 4 4 4 4 1 2 3 4 and 4 1 2 3 4 4 4 4 4.
Solution:
1 public class Solution { 2 public int findMin(int[] num) { 3 if (num.length==0) return -1; 4 int start = 0, end = num.length-1; 5 int min = findMinRecur(num,start,end); 6 return min; 7 } 8 9 //NOTE: We need to consider the ending case that start==end, this happens when there is only one element in the array! 10 public int findMinRecur(int[] num, int start, int end){ 11 if (start==end-1 || start==end){ 12 if (num[start]<num[end]) return num[start]; 13 else return num[end]; 14 } 15 16 int mid = (start+end)/2; 17 int min = Integer.MAX_VALUE; 18 if (num[mid]<num[start]){ 19 min = findMinRecur(num,start,mid); 20 return min; 21 } 22 23 if (num[mid]>num[end]){ 24 start = mid; 25 min = findMinRecur(num,mid,end); 26 return min; 27 } 28 29 if (num[mid]==num[start] && num[mid]==num[end]){ 30 int min1 = findMinRecur(num,start,mid); 31 int min2 = findMinRecur(num,mid,end); 32 if (min1<min2) return min1; 33 else return min2; 34 } 35 36 if (num[start]>num[end]) return num[end]; 37 else return num[start]; 38 } 39 }
Leetcode-Find Minimum in Rotated Sorted Array II
标签：style blog io ar color os sp for on
原文地址：http://www.cnblogs.com/lishiblog/p/4114637.html
