Given an integer n, return the least number of perfect square numbers that sum to n.
A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.
Example 1:
Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13 Output: 2 Explanation: 13 = 4 + 9.
Constraints:
1 <= n <= 104
For dynamic programming, define dp[i] to represent the least number of perfect square numbers that sum to i.
class Solution:
def numSquares(self, n: int) -> int:
dp = [0 for i in range(n + 1)]
for i in range(1, n + 1):
j, mi = 1, 0x3f3f3f3f
while j * j <= i:
mi = min(mi, dp[i - j * j])
j += 1
dp[i] = mi + 1
return dp[n]class Solution {
public int numSquares(int n) {
List<Integer> ans = new ArrayList<>();
ans.add(0);
while (ans.size() <= n) {
int m = ans.size(), val = Integer.MAX_VALUE;
for (int i = 1; i * i <= m; i++) {
val = Math.min(val, ans.get(m - i * i) + 1);
}
ans.add(val);
}
return ans.get(n);
}
}function numSquares(n: number): number {
let dp = new Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
let min = Infinity;
for (let j = 1; j * j <= i; ++j) {
min = Math.min(min, dp[i - j * j]);
}
dp[i] = min + 1;
}
return dp.pop();
};/*
* @lc app=leetcode.cn id=279 lang=golang
* 动态规划的思路，状态转移方程：dp[n] = min(dp[n-1*1]+1, dp[n-2*2]+1, ..., dp[n-k*k]+1), ( 0< k*k <=n )
*/
func numSquares(n int) int {
if n <= 0 {
return 0
}
dp := make([]int, n+1) // 多申请了一份整形，使代码更容易理解, dp[n] 就是 n 的完全平方数的求解
for i := 1; i <= n; i++ {
dp[i] = i // 初始值 dp[n] 的最大值的解，也是最容易求的解
for j := 0; j*j <= i; j++ {
dp[i] = minInt(dp[i-j*j]+1, dp[i])
}
}
return dp[n]
}
func minInt(x, y int) int {
if x < y {
return x
}
return y
}