题目描述:

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.

与上一题类似, 但是有重复元素, 所以要用遍历的方法跳过这些元素.

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class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0, right = nums.size() - 1, mid = (right + left) / 2;
        while(left < right){
            if(nums[mid] > nums[right]) left = mid + 1;
            else if(nums[mid] == nums[right]) right--;
            else right = mid;
            mid = (right + left) / 2;
        }
        return nums[left];
    }
};