Lintcode: Perfect Square

Topic

Give a positive integer n to find several complete squares (e.g. 1, 4, 9, ...) Make their sum equal to N. You need to have the fewest number of squares.

Sample Example

Give n = 12 , return 3 because 12 = 4 + 4 + 4 .
Give n = 13 , return 2 because 13 = 4 + 9 .

Solving

Title tag: Depth-First traversal

The following deep search, by finding all the results, and then find the shortest one

 Public classSolution {/**     * @paramn a positive integer *@returnAn integer*/     Public intNumsquares (intN) {//Write Your code here        intUp = (int) math.sqrt (n); ArrayList<ArrayList<Integer>> result =NewArraylist<arraylist<integer>>(); ArrayList<Integer> list =NewArraylist<integer>(); intsum = 0;        DFS (RESULT,LIST,UP,SUM,N); intmin = result.get (0). Size ();  for(intI=0;i<result.size (); i++) {min=math.min (Min,result.get (i). Size ()); }        returnmin; }     Public voidDFS (arraylist<arraylist<integer>> result,arraylist<integer> list,intStartintSumintN) {        if(n = =sum) {Result.add (NewArraylist<integer>(list)); return; }        if(Start==0 | | sum>N)return;  for(inti=start;i>=1;i--){                        if(Sum+i*i >N) {                Continue;                        } list.add (i); DFS (Result,list,i,sum+i*i,n); List.remove (List.size ()-1); }    }}

Input:

1684
时候内存溢出

不记录结果，只统计次数，改成下面的在1684时候时间超时

 Public classSolution {/**     * @paramn a positive integer *@returnAn integer*/     intMincount =Integer.max_value;  Public intNumsquares (intN) {//Write Your code here        intUp = (int) math.sqrt (n); ArrayList<ArrayList<Integer>> result =NewArraylist<arraylist<integer>>(); ArrayList<Integer> list =NewArraylist<integer>(); intsum = 0;        DFS (RESULT,LIST,UP,SUM,N); //int min = result.get (0). Size (); //for (int i=0;i<result.size (); i++) {//min = math.min (Min,result.get (i). Size ()); // }        returnMincount; }     Public voidDFS (arraylist<arraylist<integer>> result,arraylist<integer> list,intStartintSumintN) {        if(n = =sum) {            //Result.add (new arraylist<integer> (list));Mincount =math.min (Mincount,list.size ()); return; }        if(Start==0 | | sum>N)return;  for(inti=start;i>=1;i--){                        if(Sum+i*i >N) {                Continue;                        } list.add (i); DFS (Result,list,i,sum+i*i,n); List.remove (List.size ()-1); }    }}

Dynamic Programming Problem Solving

Reference links

If a number x can be represented as an arbitrary number a plus a squared number of BXB, that is, X=A+BXB, then the number of squares that can be composed of a minimum squared number plus 1 (because b*b is already a square number).

 Public classSolution {/**     * @paramn a positive integer *@returnAn integer*/     intMincount =Integer.max_value;  Public intNumsquares (intN) {//Write Your code here                    int[] DP =New int[N+1]; //The result of all non-square numbers is set to maximum, which is guaranteed to not be selected after comparisonArrays.fill (DP, integer.max_value); //Place the result of all squares 1         for(inti = 0; I * I <= N; i++) {Dp[i* I] = 1; }        //from small to large to find any number a         for(intA = 0; a <= N; a++){            //from small to large to find square number BXB             for(intb = 0; A + b * b <= n; b++){                //because A+b*b is probably the square number itself, we're going to take two of the smallerDp[a + b * b] = Math.min (Dp[a] + 1, dp[a + b *b]); }        }        returnDp[n]; }}    

Lintcode: Perfect Square