[LeetCode] 279. Perfect Squares（完全平方数）

• Difficulty: Medium

• Related Topics: Math, Dynamic Programming, Breadth-first Search

• Link: https://leetcode.com/problems/perfect-squares/

Description

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
给定一个正整数 n，现需要用完全平方数（1, 4, 9, 16）的和凑出 n，问所需平方数的最小值。

Examples

Example 1

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

Example 2

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

Solution

先给出一个 naive 的做法，暴力搜索，大致基于下式进行计算（将题目所求的函数记为 (f(n))）：

[f(n) = egin{cases} 1& n 是完全平方数\ min(f(i) + f(n - i)) ; { i in [1, n / 2] } & else end{cases} ]

为了防止重复计算，还是采用了记忆化搜索的方法，结果华丽丽的 TLE 了（input = 6175），而我本地跑这组测试数据的时候直接爆栈了……

import kotlin.math.floor
import kotlin.math.min
import kotlin.math.sqrt

class Solution {
    private val memo = hashMapOf<Int, Int>()

    fun numSquares(n: Int): Int {
        if (memo.containsKey(n)) {
            return memo.getValue(n)
        }
        if (n.isPerfectSquare()) {
            memo[n] = 1
            return 1
        }
        var result = Int.MAX_VALUE
        for (i in 1..(n / 2)) {
            result = min(result, numSquares(i) + numSquares(n - i))
        }
        memo[n] = result
        return result
    }

    private fun Int.isPerfectSquare(): Boolean {
        val root = floor(sqrt(this.toDouble())).toInt()
        return root * root == this
    }
}

自然只能找其它解法了，以下一个解法来自 discussion。考虑以下等式（还是一样，将原函数记为 (f(n))）：

[f(0) = 0 \ f(1) = f(0) + 1 = 1 \ f(2) = f(1) + 1 = 2 \ f(3) = f(2) + 1 = 3 \ f(4) = min(f(4 - 1 imes 1) + 1, f(4 - 2 imes 2) + 1) = min(f(3) + 1, f(0) + 1) = 1 \ f(5) = min(f(5 - 1 imes 1) + 1, f(5 - 2 imes 2) + 1) = min(f(4) + 1, f(1) + 1) = 2 \ ... \ f(13) = min(f(13 - 1 imes 1) + 1, f(13 - 2 imes 2) + 1, f(13 - 3 imes 3) + 1) = min(f(12) + 1, f(9) + 1, f(4) + 1) = 2 ]

所以最后的状态转移方程是：

[f(n) = min(f(n - i imes i) + 1)（其中 n - i imes i geq 0 且 i geq 1） ]

代码如下：

import kotlin.math.min

class Solution {
    fun numSquares(n: Int): Int {
        val dp = IntArray(n + 1) { Int.MAX_VALUE }
        dp[0] = 0
        for (i in 1..n) {
            var min = Int.MAX_VALUE
            var j = 1
            while (i - j * j >= 0) {
                min = min(min, dp[i - j * j] + 1)
                j++
            }
            dp[i] = min
        }
        return dp.last()
    }
}