Find Minimum in Rotated Sorted Array II

Question: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

题目：

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

 

思路：

两端寻找，跳过重复项，直至 nums[start] 与前项不同，nums[end] 与后项不同，且nums[start] 与 nums[end] 不相同。

其他部分，同 Find Minimum in Rotated Sorted Array。

 

public int findMin(int[] nums) {
    if(nums == null || nums.length == 0) {
        return 0;
    }
    
    int start = 0;
    int end = nums.length - 1;
    int minNum = nums[start];
    
    while(start < end) {
        int startNum = nums[start++];
        int endNum = nums[end--];
        minNum = Math.min(minNum, Math.min(startNum, endNum));
        
        while(start < end && nums[start] == startNum) {
            start++;
        }
        
        while(end > start && nums[end] == endNum) {
            end--;
        }
        
        if(start == end) {
            return Math.min(minNum, nums[start]);
        } else if(nums[start] == nums[end]) {
            continue;
        }
        
        int mid = (start + end)/2;　　　　// 此处 nums[mid],nums[start],nums[end]的值将均不同
        if(nums[start] < nums[end]) {
            return Math.min(minNum, nums[start]);
        } else if(nums[end] < nums[mid]) {
            start = mid + 1;
            minNum = Math.min(minNum, nums[end]);
        } else if(nums[start] > nums[mid]) {
            start++;
            end = mid;
            minNum = Math.min(minNum, nums[mid]);
        } 
    }
    
    return minNum;
}