Skip to content

Latest commit

 

History

History
297 lines (250 loc) · 7.04 KB

File metadata and controls

297 lines (250 loc) · 7.04 KB

English Version

题目描述

给你一个整数 n ,返回 和为 n 的完全平方数的最少数量

完全平方数 是一个整数,其值等于另一个整数的平方;换句话说,其值等于一个整数自乘的积。例如,14916 都是完全平方数,而 311 不是。

 

示例 1:

输入:n = 12
输出:3 
解释:12 = 4 + 4 + 4

示例 2:

输入:n = 13
输出:2
解释:13 = 4 + 9

 

提示:

  • 1 <= n <= 104

解法

方法一:动态规划(完全背包)

我们定义 $f[i][j]$ 表示使用数字 $1, 2, \cdots, i$ 的完全平方数组成和为 $j$ 的最少数量。初始时 $f[0][0] = 0$,其余位置的值均为正无穷。

我们可以枚举使用的最后一个数字的数量 $k$,那么:

$$ f[i][j] = \min(f[i - 1][j], f[i - 1][j - i^2] + 1, \cdots, f[i - 1][j - k \times i^2] + k) $$

其中 $i^2$ 表示最后一个数字 $i$ 的完全平方数。

不妨令 $j = j - i^2$,那么有:

$$ f[i][j - i^2] = \min(f[i - 1][j - i^2], f[i - 1][j - 2 \times i^2] + 1, \cdots, f[i - 1][j - k \times i^2] + k - 1) $$

将二式代入一式,我们可以得到以下状态转移方程:

$$ f[i][j] = \min(f[i - 1][j], f[i][j - i^2] + 1) $$

最后答案即为 $f[m][n]$

时间复杂度 $O(m \times n)$,空间复杂度 $O(m \times n)$。其中 $m$$sqrt(n)$ 的整数部分。

注意到 $f[i][j]$ 只与 $f[i - 1][j]$$f[i][j - i^2]$ 有关,因此我们可以将二维数组优化为一维数组,空间复杂度降为 $O(n)$

相似题目:

class Solution:
    def numSquares(self, n: int) -> int:
        m = int(sqrt(n))
        f = [[inf] * (n + 1) for _ in range(m + 1)]
        f[0][0] = 0
        for i in range(1, m + 1):
            for j in range(n + 1):
                f[i][j] = f[i - 1][j]
                if j >= i * i:
                    f[i][j] = min(f[i][j], f[i][j - i * i] + 1)
        return f[m][n]
class Solution {
    public int numSquares(int n) {
        int m = (int) Math.sqrt(n);
        int[][] f = new int[m + 1][n + 1];
        for (var g : f) {
            Arrays.fill(g, 1 << 30);
        }
        f[0][0] = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= i * i) {
                    f[i][j] = Math.min(f[i][j], f[i][j - i * i] + 1);
                }
            }
        }
        return f[m][n];
    }
}
class Solution {
public:
    int numSquares(int n) {
        int m = sqrt(n);
        int f[m + 1][n + 1];
        memset(f, 0x3f, sizeof(f));
        f[0][0] = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= i * i) {
                    f[i][j] = min(f[i][j], f[i][j - i * i] + 1);
                }
            }
        }
        return f[m][n];
    }
};
func numSquares(n int) int {
	m := int(math.Sqrt(float64(n)))
	f := make([][]int, m+1)
	const inf = 1 << 30
	for i := range f {
		f[i] = make([]int, n+1)
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= m; i++ {
		for j := 0; j <= n; j++ {
			f[i][j] = f[i-1][j]
			if j >= i*i {
				f[i][j] = min(f[i][j], f[i][j-i*i]+1)
			}
		}
	}
	return f[m][n]
}
function numSquares(n: number): number {
    const m = Math.floor(Math.sqrt(n));
    const f: number[][] = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(1 << 30));
    f[0][0] = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = 0; j <= n; ++j) {
            f[i][j] = f[i - 1][j];
            if (j >= i * i) {
                f[i][j] = Math.min(f[i][j], f[i][j - i * i] + 1);
            }
        }
    }
    return f[m][n];
}
impl Solution {
    pub fn num_squares(n: i32) -> i32 {
        let (row, col) = ((n as f32).sqrt().floor() as usize, n as usize);
        let mut dp = vec![vec![i32::MAX; col + 1]; row + 1];
        dp[0][0] = 0;
        for i in 1..=row {
            for j in 0..=col {
                dp[i][j] = dp[i - 1][j];
                if j >= i * i {
                    dp[i][j] = std::cmp::min(dp[i][j], dp[i][j - i * i] + 1);
                }
            }
        }
        dp[row][col]
    }
}

方法二

class Solution:
    def numSquares(self, n: int) -> int:
        m = int(sqrt(n))
        f = [0] + [inf] * n
        for i in range(1, m + 1):
            for j in range(i * i, n + 1):
                f[j] = min(f[j], f[j - i * i] + 1)
        return f[n]
class Solution {
    public int numSquares(int n) {
        int m = (int) Math.sqrt(n);
        int[] f = new int[n + 1];
        Arrays.fill(f, 1 << 30);
        f[0] = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = i * i; j <= n; ++j) {
                f[j] = Math.min(f[j], f[j - i * i] + 1);
            }
        }
        return f[n];
    }
}
class Solution {
public:
    int numSquares(int n) {
        int m = sqrt(n);
        int f[n + 1];
        memset(f, 0x3f, sizeof(f));
        f[0] = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = i * i; j <= n; ++j) {
                f[j] = min(f[j], f[j - i * i] + 1);
            }
        }
        return f[n];
    }
};
func numSquares(n int) int {
	m := int(math.Sqrt(float64(n)))
	f := make([]int, n+1)
	for i := range f {
		f[i] = 1 << 30
	}
	f[0] = 0
	for i := 1; i <= m; i++ {
		for j := i * i; j <= n; j++ {
			f[j] = min(f[j], f[j-i*i]+1)
		}
	}
	return f[n]
}
function numSquares(n: number): number {
    const m = Math.floor(Math.sqrt(n));
    const f: number[] = Array(n + 1).fill(1 << 30);
    f[0] = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = i * i; j <= n; ++j) {
            f[j] = Math.min(f[j], f[j - i * i] + 1);
        }
    }
    return f[n];
}
impl Solution {
    pub fn num_squares(n: i32) -> i32 {
        let (row, col) = ((n as f32).sqrt().floor() as usize, n as usize);
        let mut dp = vec![i32::MAX; col + 1];
        dp[0] = 0;
        for i in 1..=row {
            for j in i * i..=col {
                dp[j] = std::cmp::min(dp[j], dp[j - i * i] + 1);
            }
        }
        dp[col]
    }
}