Lintcode: Finding the minimum value in a rotated sorted array II

look for the minimum value in the rotated sorted array II

Suppose a rotated sorted array whose starting position is unknown (e.g. 0 1 2 4 5 6 7 May become 4 56 7 0 1 2).

You need to find the smallest of these elements.

There may be duplicate elements in the array.

Solving

Violence Direct Linear Lookup

Alternatively, linear finds the number corresponding to the first starting descending position

The dichotomy should be considered

Recursion + two points

 Public classSolution {/**     * @paramnum:a rotated sorted array *@return: The minimum number in the array*/     Public intFindmin (int[] num) {        //Write your code here        if(num = =NULL|| Num.length = = 0)            return0; if(Num.length = = 1)            returnNum[0]; returnFindmin (num,0,num.length-1); }     Public intFindmin (int[] num,intLeftintRight ) {        if(left = =Right )returnNum[left]; if(right = = left + 1)            returnmath.min (Num[left],num[right]); intMid = (left + right)/2; if(Num[right] >Num[left])returnNum[left]; Else if(Num[right] = =num [left])returnFindmin (Num,left + 1, right); Else if(Num[mid] >=Num[left])returnfindmin (num,mid,right); Else             returnfindmin (Num,left,mid); }}

Java Code

Two points

 Public classSolution {/**     * @paramnum:a rotated sorted array *@return: The minimum number in the array*/     Public intFindmin (int[] num) {        //Write your code here        if(num = =NULL|| Num.length = = 0)            return0; if(Num.length = = 1)            returnNum[0]; intLow = 0; intHigh = Num.length-1; intMID =Low ;  while(Low <High ) {Mid= (low + high)/2; if(Num[mid] >Num[high]) { Low= Mid + 1; }Else if(Num[mid] <Num[high]) { High=mid; }Else{ High--; }        }        returnNum[low]; } }

Java Code

Lintcode: Finding the minimum value in a rotated sorted array II