"Leetcode" Search in rotated Sorted Array (hard)

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 ).

You is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

Idea: Two-point search, remove half of the error options at a time.

Note that every time L = m + 1, r = m-1 prevents infinite loops.

intSearchintA[],intNinttarget) {    intL =0, R = N-1;  while(L <=r)//note that there is an equal sign {intm = (L + r)/2; if(A[m] = =target)returnm; if(A[l] <= a[m] && a[m] <= A[r])//Order of        {            if(A[m] >target) R= M-1; ElseL= m +1; }        Else if(A[l] >= a[m] && a[m] <= A[r])//the beginning goes to the left half.        {            if(A[m] < target && target <= A[r])//in the right-hand partL = m +1; ElseR= M-1; }        Else //the beginning goes to the right half.        {            if(A[l] <= target && target <= a[m])//in the left half partr = M-1; ElseL= m +1; }    }    return-1;}

Great God simplified version: the idea of removing half of the options is different

intSearchintA[],intNinttarget) {    intLo =0; intHi = N-1;  while(Lo <=hi) {        intMid = (lo + hi)/2; if(A[mid] = = target)returnmid; if(A[lo] <=A[mid]) {            if(Target >= A[lo] && Target <A[mid]) {Hi= Mid-1; } Else{lo= Mid +1; }        } Else {            if(Target > A[mid] && target <=A[hi]) {Lo= Mid +1; } Else{Hi= Mid-1; }        }    }    return-1;}

"Leetcode" Search in rotated Sorted Array (hard)