Find Peak Element

Find Peak Element

Problem:

A peak element is an element, which is greater than its neighbors.

Given an input array where num[i] ≠ num[i+1] , the find a peak element and return its index.

The array may be contain multiple peaks, in this case return the index to any one of the peaks is fine.

May imagine num[-1] = num[n] = -∞ .

For example, in array [1, 2, 3, 1] , 3 is a peak element and your function should return the index number 2.

Ideas:

Two-point Search

My train of thought:

 Public classSolution { Public intFindpeakelement (int[] num) {        if(num = =NULL|| Num.length = = 0)return-1; intLen =num.length; if(len = 1 | | num[0] > NUM[1])return0; if(Num[len-1] > Num[len-2])returnLen-1; intleft = 1; intright = Len-2; returngetpeakindex (num, left, right); }     Public intGetpeakindex (int[] num,intLeftintRight ) {        if(Left > right)return-1; if(left = =Right ) {            if(Num[left] > Num[left-1] && num[left] > num[left + 1])returnLeft ; Else return-1; }        intMid = (left + right)/2; if(Num[mid] > Num[mid-1] && num[mid] > num[mid + 1])returnmid; intindex = Getpeakindex (num, left, mid-1); if(Index = =-1) Index= Getpeakindex (num, mid + 1, right); returnindex; }        }

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Improved post-algorithm:

 Public classSolution { Public intFindpeakelement (int[] num) {        if(num = =NULL|| Num.length = = 0)return-1; intLen =num.length; if(len = = 1)return0; intleft = 0; intright = Len-1;  while(Left <=Right ) {            intMid = (left + right)/2; if(Mid = = 0)            {                if(Num[mid] > Num[mid+1])returnmid; }            Else if(Mid = = Len-1)            {                if(Num[mid] > Num[mid-1])returnmid; }            Else            {                if(Num[mid] > Num[mid-1] && num[mid] > num[mid+1])returnmid; }            if(Num[mid] < num[mid+1]) left = mid + 1; Elseright =mid; }        return-1; }}

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Others code:

classSolution { Public:    intFindpeakelement (Constvector<int> &num) {        intLeft=0,right=num.size ()-1;  while(left<=Right ) {            if(left==Right )returnLeft ; intMid= (left+right)/2; if(num[mid]<num[mid+1]) left=mid+1; Else Right=mid; }    }};

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The Learning Place:

• First of all there must be peak points, because the leftmost num[-1]=num[n]= negative infinity, there must be a peak
• If the middle point is not peak, move towards the left and right sides of the rising
• Left and right, are going up the place to go, meet the place is peak point (this is a bit difficult to understand AH), did not prove it, but a thought is such a thing
• Because the peak point is definitely there, right = mid does not cause a dead loop and always jumps out of the loop.

Find Peak Element