[LintCode/LeetCode] Perfect Squares

Problem

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

Example

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

Note

这道题在OJ有很多解法，公式法，递归法，动规法，其中公式法时间复杂度最优（four square theorem）。
不过我觉得考点还是在动规吧，也更好理解。

Solution

1. 动态规划法

public class Solution {
    public int numSquares(int n) {
        //建立空数组dp：从0到n每个数包含最少平方数情况，先fill所有值为Integer.MAX_VALUE;
        int[] dp = new int[n+1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        //将0到n范围内所有平方数的dp值赋1；
        for (int i = 0; i*i <= n; i++) {
            dp[i*i] = 1;
        }
        //两次循环更新dp[i+j*j]，当它本身为平方数时，dp[i+j*j] < dp[i]+1
        for (int i = 0; i <= n; i++) {
            for (int j = 0; i+j*j <= n; j++) {
                dp[i+j*j] = Math.min(dp[i]+1, dp[i+j*j]);
            }
        }
        return dp[n];
    }
}

2. 简化动态规划法

public class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n+1];
        for (int i = 0; i <= n; i++) {
            dp[i] = i;
            for (int j = 0; j*j <= i; j++) {
                dp[i] = Math.min(dp[i], dp[i-j*j]+1);
            }
        }
        return dp[n];
    }
}

3. 四平方和定理法

public class Solution {
    public int numSquares (int n) {
        int m = n;
        while (m % 4 == 0)
            m = m >> 2;
        if (m % 8 == 7)
            return 4;
        int sqrtOfn = (int) Math.sqrt(n);
        if (sqrtOfn * sqrtOfn == n) //Is it a Perfect square?
            return 1;
        else {
                for (int i = 1; i <= sqrtOfn; ++i){
                    int remainder = n - i*i;
                    int sqrtOfNum = (int) Math.sqrt(remainder);
                    if (sqrtOfNum * sqrtOfNum == remainder)
                        return 2;
                }
            }
        return 3;
    }
}