[Leetcode]Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example,1, 4, 9, 16, ...) which sum ton.

For example, given n = 12, return 3 because12 = 4 + 4 + 4; givenn =13, return2 because13 = 4 + 9.

class Solution {public:    /*algorithm: dfs solution        consturct an array (1,...,sqrt(n))        enumerate array elements sum to n    */    void dfs(int start,int cur,int cnt,int &minCnt){        if(cur < 0)return;        if(cur == 0){            minCnt = min(minCnt,cnt);            return;        }        for(int i = start;i > 0;i--){            if(cnt >= minCnt)break;            if(cur < (i*i))continue;            dfs(i,cur - i*i,cnt+1,minCnt);        }    }    int numSquares(int n) {        int d = sqrt(n);//12==>3        int minCnt = INT_MAX;        dfs(d,n,0,minCnt);        return minCnt;    }};

class Solution {public:    /*alorithm Lagrange's four-square theorem    time O(sqrt(n))   */    int numSquares(int n) {        while(n%4 == 0) n /= 4;        if(n%8 == 7)return 4;        for(int i = 0;i*i <= n;i++){            int j = sqrt(n-i*i);            if((i*i + j*j) == n)                return !!i + !!j;        }        return 3;    }};

class Solution {public:    /*algorithm: dp solution        1)get d = sqrt(n)        1)for n          f(n) = min{f(n-i*i) + f(i*i),f(n)},among  f(i*i) = 1 (0<i <= d)          f(0) = 0,          f(1) = 1          time O(n*sqrt(n)) space O(n)    */    int numSquares(int n) {        vector<int>dp(n+1,INT_MAX);//[0,n+1)        dp[0] = 0,dp[1] = 1;        int k = sqrt(n);        for(int i = 1;i < k+1;i++)dp[i*i] = 1;                for(int i = 2;i < n+1;i++){            int k = sqrt(i);            for(int j = 1;j <= k;j++){                dp[i] = min(dp[j*j] + dp[i-j*j],dp[i]);            }        }        return dp[n];    }};