HAPPYLEETCODE64:SQRT (x)

Implement int sqrt(int x) .

This problem is essentially the largest integer under sqrt (x). Binary search is a more easily thought-out approach. Another, on the internet and learn the other people's Newton iterative method.

This is my original writing, the write is wrong, the complexity is too high

classSolution { Public:    intsqrtintx) {if(x <=0)            return 0; if(x = =0|| x = =1)            returnx; Long LongStart =1; Long Longend = x >>1; intindex =start;  while(Start <=end) {Index= (start + end) >>1; if(Index*index = =x)returnindex; Else{                if(End*end <=x)returnend; if(End-start = =1)                    returnstart; if(Index*index >x) End= Index-1; if(Index*index <x) Start=index; }        }        returnindex; }};

In fact, the idea of binary search is right, but in some small details in Shanghai need to pay particular attention.

The standard code can be written like this:

classSolution { Public:    intsqrtintx) {Long Longi =0; Long Longj = x/2+1;  while(I <=j) {Long LongMid = (i + j)/2; Long LongSq = Mid *mid; if(Sq = =x)returnmid; Else if(Sq <x) I= Mid +1; ElseJ= Mid-1; }        returnJ; }};

The problem of overflow is considered perfectly. This is the place where I should study hard.

Newton's iterative FA-solving

With this method, you can get an iterative formula Xi+1=xi-(XI2-N)/(2XI) = XI-XI/2 + N/(2xi) = XI/2 + N/2xi = (xi + n/xi)/2.

Then there is the following code:

 int  sqrt (int   x) { if  (x = =  0 ) return  0     ;  double  last = 0  ;  double  res = 1  ;   while  (res!= = res;    Res  = (res + x/res)/2  ;  return  int   

Solve the problem of double, can refer to the above code

DoublesqrtDoublex) {if(x = =0)return 0; DoubleLast =0.0; Doubleres =1.0;  while(!Euqal (Res,last)) { Last=Res; Res= (res + x/res)/2; }    returnRes;}BOOLEuqal (DoubleNUM1,Doublenum2) {    if((NUM1-NUM2) <0.0000001&& (num1-num2) >-0.0000001)        return true; Else        return false;}

Reference:

Http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html

HAPPYLEETCODE64:SQRT (x)