Leetcode. Find Minimum in rotated Sorted Array

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.

Array A,

1) If A[MID] < A[low], indicate minimum between [mid, Low]. (Presence rotated)

2) if A[MID] > A[low], there are two cases:

A) A[mid] > A[high], minimum between [Mid, High]. (Presence rotated)

b) A[mid] < A[high], minimum between [Low, mid]. (Rotated not present)

Notice termination condition: A[mid] Less than left or right side or low = = Mid

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(Mid = =0|| Num[mid] < Num[mid-1]) && (Mid = = Num.size ()-1|| Num[mid] < Num[mid +1]))8                  Break;9             if(Num[mid] <Num[low])Ten             { OneHigh = mid-1; A             } -             Else if(Num[mid] >Num[low]) -             { the                 if(Num[mid] >Num[high]) -Low = mid +1; -                 Else if(Num[mid] <Num[high]) -High = mid-1; +             } -             Else +             { A                 returnmin (Num[low], Num[high]); at             } -         } -          -         returnNum[mid]; -}

Leetcode. Find Minimum in rotated Sorted Array