Leetcode Perfect Squares

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Problem

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

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

Solution 1 - pure recursion

求n 的perfect square, 可以化简成求 n 减去若干个 i*i之后的数的perfect square.

class Solution {public:    int numSquares(int n) {        if(n <= 1) return n;                int rst = n;        for( int i = 2; i*i <= n; i++){            int a = n / (i * i), b = n % (i * i);            rst = min (rst, numSquares(b) + a);        }        return rst;    }};

Solution 2 - recursion with memory

用一个数字，纪录上面的过程中重复值。

class Solution {    int helper( int n, vector<int>& arr){        if(arr[n] != -1) return arr[n];                int rst = n;        for( int i = 2; i*i <= n; i++){            int a = n / (i * i), b = n % (i * i);            rst = min (rst, helper(b, arr) + a);        }        arr[n] = rst;        return rst;                   }public:    int numSquares(int n) {        if(n <= 1) return n;        vector<int> arr(n+1, -1);        helper(n, arr);        return arr[n];    }};

Solution 3 - DP

其实上面的解法二已经很接近DP了

dp[ x + y*y ] = min( dp[ x+ y*y ] , dp[ x ] + 1 );

class Solution {public:    int numSquares(int n) {        if(n <= 1) return n;        vector<int> dp(n+1, INT_MAX);                      for (int i = 1; i * i <= n; i++) {            dp[i * i] = 1;        }        for (int i = 1; i <= n; i++) {            for (int j = 1; i + j * j <= n; j++) {                dp[i + j * j] = min(dp[i] + 1, dp[i + j * j]);            }        }        return dp[n];    }};

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