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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.
思路分析:这题类似于LeetCode Find Minimum in Rotated Sorted Array ,具体解析可以参考前一篇,这题的特殊之处是数组中可以有重复元素,那么当A[m]=A[l]时,执行l++即可。此时,就只可以排除掉一个元素,而不是一半元素,最坏情况下算法时间复杂度是O(n),不再是O(logN)。
AC Code:
public class Solution {
public int findMin(int[] nums) {
int n = nums.length;
int l = 0;
int r = n-1;
int min = nums[0];
while(l < r){
int m = l + (r-l) / 2;
if(nums[m] < nums[l]) {
min = Math.min(nums[m], min);
r = m;
} else if(nums[m] > nums[l]){
min = Math.min(nums[l], min);
l = m;
} else {
l++;
}
}
min = Math.min(nums[l], min);
min = Math.min(nums[r], min);
return min;
}
}