0279. Perfect Squares (M)

Perfect Squares (M)

题目

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

Example 1:

Input: n = 12
Output: 3 
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

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题意

将一个整数转换为x个完全平方数的和，求x的最小值。

思路

直接使用递归会超时，使用记忆化搜索进行优化。

也可以使用动态规划：dp[i]表示整数i最少由dp[i]个完全平方数相加得到，dp[i]初始值为1(自己就是完全平方数)或i(由i个1相加得到)，递推公式为：(dp[i]=Math.min(dp[i], dp[j] + dp[i-j]))。

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代码实现

Java

记忆化搜索

class Solution {
    public int numSquares(int n) {
        int[] record = new int[n + 1];
        Arrays.fill(record, -1);
        return dfs(n, record);
    }

    private int dfs(int n, int[] record) {
        if (n <= 1) {
            return n;
        }

        if (record[n] != -1) {
            return record[n];
        }

        int min = n;
        for (int i = 1; i * i <= n; i++) {
            min = Math.min(min, dfs(n - i * i, record));
        }

        record[n] = min + 1;
        return min + 1;
    }
}

动态规划

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            int root = (int) Math.sqrt(i);
            dp[i] = root * root == i ? 1 : i;
            for (int j = 1; j <= i / 2; j++) {
                dp[i] = Math.min(dp[i], dp[i - j] + dp[j]);
            }
        }
        return dp[n];
    }
}