[LeetCode] Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example,

1, 4, 9, 16, ...

) which sum to n.

For example, given n =

12

, return

3

because

12 = 4 + 4 + 4

; given n =

13

, return

2

because

13 = 4 + 9

.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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解题思路

动态规划法。初始化时令

dp[i * i] = 1

，状态转移方程为

dp[i + j * j] = min(dp[i] + 1, dp[i + j * j])

;

实现代码

C++：

// Runtime: 544 ms
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1, 0x7fffffff);
for (int i = 0; i * i <= n; i++)
{
dp[i * i] = 1;
}

for (int i = 1; i <= n; i++)
{
for (int j = 1; i + j * j <= n; j++)
{
dp[i + j * j] = min(dp[i] + 1, dp[i + j * j]);
}
}

return dp
;
}
};

Java：

// Runtime: 69 ms
public class Solution {
public int numSquares(int n) {
int dp[] = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
for (int i = 1; i * i <= n; i++) {
dp[i * i] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; i + j * j <= n; j++) {
dp[i + j * j] = Math.min(dp[i] + 1, dp[i + j * j]);
}
}

return dp
;
}
}