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package Algorithms.binarySearch;

/*

* Find Minimum in Rotated Sorted Array II Total Accepted: 2541 Total Submissions: 9558 My Submissions Question Solution

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 FindMin2 {

public int findMin(int[] num) {

if (num == null || num.length == 0) {

return 0;

}

int len = num.length;

if (len == 1) {

return num[0];

} else if (len == 2) {

return Math.min(num[0], num[1]);

}

int left = 0;

int right = len - 1;

while (left < right - 1) {

int mid = left + (right - left) / 2;

// In this case, the array is sorted.

// 这一句很重要，因为我们移除一些元素后，可能会使整个数组变得有序...

if (num[left] < num[right]) {

return num[left];

}

// left side is sorted. CUT the left side.

if (num[mid] > num[left]) {

left = mid;

// left side is unsorted, right side is sorted. CUT the right side.

} else if (num[mid] < num[left]) {

right = mid;

} else {

left++;

}

}

return Math.min(num[left], num[right]);

}

}