字符串算法——查找有序数组旋转后最小值(有重复元素)(Find Minimum in Rotated Sorted Array II)

问题:

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 an array sorted in ascending order 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.

思路一：同无重复元素数组一样，采用最简单的直接遍历，依次比较，不过这里时间复杂度不满足条件。

class Solution {
public int findMin(int[] nums) {
int len = nums.length;
int minIndex = 0;
for(int i = 1;i<len;i++){
if(nums[minIndex]>nums[i]){
minIndex = i;
}
}
return nums[minIndex]；
}
}

思路二：同无重复元素数组一样，可以采用二分查找的算法，但由于存在重复元素，所以无法确定最小值是落在中间索引的哪一侧，对两侧分别查找，比较两侧查找的最小值。

class Solution {
public int findMin(int[] nums) {
int len = nums.length;
return dfs(nums,0,len-1);
}
private int dfs(int[]nums,int min,int max){
if(min==max){
return nums[min];
}
int mid = (min+max)/2;//中间索引值
int left = Integer.MAX_VALUE;
int right = Integer.MAX_VALUE;
if(nums[mid]>=nums[max]){
left = dfs(nums,mid+1,max);//可能落在右侧
}
if(nums[mid]<=nums[max]){//可能落在左侧
right = dfs(nums,min,mid);
}
return Math.min(left,right);//比较两侧查找出的值
}
}