Leetcode|Find Minimum in Rotated Sorted Array[二分查找]

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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.

You may assume no duplicate exists in the array.

参考leetcode solution;
对于数组A,AL代表最左边的元素,AR代表最右边的元素;
如果没旋转,AL

int findMin(vector<int>& nums) {
        int n=nums.size();
        if(n==1) return nums[0];

        int first=0,last=n-1;
        //consider special conditions
        //if(nums[first]<nums[last]) return nums[first];
        while(first<last){
            int mid=(first+last)/2;
            if(nums[last]<nums[mid]) first=mid+1;
            else last=mid;
        }
        return nums[last];
    }

如果有重复元素呢?
最坏情况时间复杂度O(n);
直接去除数组初始的重复元素(nums[first]==nums[last]),后面和没重复的一样了!见下一篇博客分析!

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转载自blog.csdn.net/mike_learns_to_rock/article/details/47188249
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