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[n];
    }
};

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[n];
    }
}