Leetcode 154/153. Find Minimum in rotated Sorted Array && II

Suppose a sorted array is rotated on 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.

Assume no duplicate exists in the array.

Analysis:

For arrays that do not contain duplicate elements a[0, ..., n-1], if A[MID] < A[high], indicate the minimum value between A[low, mid].

Otherwise, between A[mid + 1, high].

1 intFindmin (vector<int> &num)2     {3         intLow =0, high = Num.size ()-1, Mid =0;4         5          while(Low <High )6         {7Mid = (low + high)/2;8             if(Num[mid] <Num[high])9High =mid;Ten             Else  OneLow = mid +1; A         } -          -         returnNum[low]; the}

If A contains duplicate elements, you need to add the judgment a[mid] = = A[high], at this time cannot use two points, such as [1, 1, 0, 1] and [0, 1, 1, 1], the smallest element

Can be at A[mid, high], or A[low, mid], at which point the range can be reduced by high--,.

1 intFindmin (vector<int> &num)2     {3         intLow =0, high = Num.size ()-1, Mid =0;4          while(Low <High )5         {6Mid = (low + high)/2;7             if(Num[mid] = =Num[high])8high--;9             Else if(Num[mid] <Num[high])TenHigh =mid; One             Else ALow = mid +1; -         } -          the         returnNum[low]; -}

Leetcode 154/153. Find Minimum in rotated Sorted Array && II