LeetCode 154. Find Minimum in Rotated Sorted Array II (Java版; Hard)

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LeetCode 154. Find Minimum in Rotated Sorted Array II (Java版; Hard)

题目描述

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.

Example 1:

Input: [1,3,5]
Output: 1
Example 2:

Input: [2,2,2,0,1]
Output: 0
Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?

第一次做; 核心: 1)让left和right最终指向同一个元素, 这样才能触发循环终止条件left==right, 如果像之前那样让left指向前部分的末尾,让right指向后部分的开头,就得修改循环终止条件; 2)必须让nums[mid]跟nums[right]比, 否则不能通过绊脚案例{10,1,10,10,10}

/*
绊脚案例{10,1,10,10,10}, 核心: 谁跟谁比? 必须让nums[mid]跟nums[right]比, 否则不能通过绊脚案例
*/
class Solution {
    public int findMin(int[] nums) {
        int n = nums.length;
        if(nums[0]<nums[n-1])
            return nums[0];
        //
        int left=0, right=n-1;
        //left==right时,循环终止, 说明left最终指向最小值
        while(left<right){
            int mid = left + ((right-left)>>1);
            //在前部分
            if(nums[mid]>nums[right])
                left = mid+1;
            //在后部分
            else if(nums[mid]<nums[right])
                right = mid;
            //nums[mid]==nums[right], 此时不知道mid在前部分还是后部分, 例如{10,1,10,10,10}
            else{
                //为了能够返回正确的索引(这里算是个小优化), 可以扩充一下else, 从而保证right--之前,nums[right-1]<=nums[right]
                if(nums[right-1]>nums[right]){
                    left=right;
                    break;
                }
                right--;
            }
        }
        return nums[left];
    }
}

LeetCode优秀题解; 上面的解法确实能返回最小值, 但是不能保证最小值的索引是正确的, 比如{1,1,1,3,1}, 最后的right是0, 并不是4,

his code is correct to return the minimum value of the array. But in terms of "find the minimum value index" it is not right.
Consider this case: 1 1 1 1 1 1 1 1 2 1 1
the min index returned is 0, while actually it should be 9.
For this case: 2 2 2 2 2 2 2 2 1 2 2
it will return the correct index, which is 8.

The reason is, the pivot index will be passed by at hi--. To avoid this, we can add the following judgement:

为了能够返回正确的索引, 可以扩充一下else, 从而保证right–之前,nums[right-1]<=nums[right]

else {
    if (nums[hi - 1] > nums[hi]) {
        lo = hi;
        break;
    }
    hi--;
}