解题报告-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.

Example 1:

Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:

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

解题思路:

第一种方案: 二维dp

和为n的数，要么以1开头，要么以4开头，要么以9开头，依次类推，本题就是求这些可能方案中，需要平方数个数最少的那组解

dp[k][n] - 表示以k开头的平方数序列，和为n时，最少需要的平方数个数

进一步递推可得

dp[k][n] = 1 + min{dp[k][n - k], dp[nextK][n - k], ...}

因为，取k是一步，取完k之后，问题变为求解和为n - k的最优解, 我们可以继续取k, 即dp[k][n - k], 或者我们直接取下一个k, dp[nextK][n - k], ....,

最终找到和为n - k时的最优解, 对解加1，即为和为n时，以k开头的最优解.

最终，我们遍历dp[i][n] i = 1, 4, 9, ..., n ，即可以得到和为n的解.

空间复杂度o(n^2) 时间复杂度o(n^3)

空间上我们可以进一步优化，不要从1开始存储到n，优化为只存储平方数

java代码

class Solution {
    //for example
    //n n - 1 n - 2
    //dp[n][num] = 1 + mim{dp[n][num - n], dp[n - 1][num - n], dp[n - 2][num - n], ..., dp[1][n]}
    //ans = min {dp[k][sum]} k = 1,2,3,...,n-1,n
    public int numSquares(int n) {
        int ans = Integer.MAX_VALUE;
        List<Integer> data = new ArrayList<>();
        int i = 0;
        for (; i * i <= n; i++) {
            data.add(i * i);
        }
        if (i * i == n) return 1;

        int cnt = data.size();
        int[][] dp = new int[cnt][1 + n];
        for (int l = 1; l < cnt; l++) {
            for (int p = 1; p <= n; p++) {
                dp[l][p] = Integer.MAX_VALUE;
            }
        }

        for (int k = 1; k < cnt; k++) {
            int val = data.get(k);
            for (int sum = val; sum <= n; sum++) {
                for (int j = k; j >= 1; j--) {
                    if (dp[j][sum - val] == Integer.MAX_VALUE) continue;
                    //System.out.println("k={}" + k + " sum=" + sum + " j={}" + j + " remain={}" + (sum - k));
                    dp[k][sum] = Math.min(dp[k][sum], 1 + dp[j][sum - val]);
                    //System.out.println("dp[k][sum]=" + dp[k][sum]);
                }
            }

            if (ans > dp[k][n]) {
                //System.out.println("dp[k][n]=" + dp[k][n]);
                ans = dp[k][n];
            }
        }

        return ans;
    }

    public boolean isSquare(int d) {
        int i = 1;
        for (; i * i < d; i++) {
        }

        if (i * i == d) return true;
        return false;
    }
}

c++代码

class Solution {
public:
    //dp[i][j] = 1 + min{dp[i][j - i], dp[nextI][j - i], ..., dp[k][j]}
    //dp[i][i] = 1 + min{dp[0][0]} = 1
    //dp[i][j] = 1 + min{dp[i][j - i], dp[ni][j - i], ..., dp[nii][j - i]}
    //if i > j, dp[i][j] = INT_MAX;
    //if i == j, dp[i][j] = 1;
    //if i < j,
    //
    int numSquares(int n) {
        int i = 1;
        vector<int> data;
        while (i * i <= n) {
            data.push_back(i * i);
            ++i;
        }

        int len = data.size();
        int dp[1 + len][1 + n];
        memset(dp, 0, sizeof(dp));
        int ans = INT_MAX;
        //cout<<"len="<<len<<endl;

        for (int i = 1; i <= len; i++) {
            //cout<<"num="<<data[i - 1]<<endl;
            for (int j = data[i - 1] - 1; j >= 1; j--) {
                dp[i][j] = INT_MAX;
                //output(i, j, dp[i][j]);
            }
        }

        for (int j = 1; j <= n; j++) {
            for (int i = 1; i <= len && data[i - 1] <= j; i++) {
                int tmp = INT_MAX;
                for (int k = i; k <= len; k++) {
                    tmp = min(tmp, dp[k][j - data[i - 1]]);
                }

                //output(i, j, tmp);

                if (tmp != INT_MAX) {
                    dp[i][j] = 1 + tmp;
                } else {
                    dp[i][j] = tmp;
                }

                //output(i, j, dp[i][j]);
            }
        }

        for (int i = 1; i <= len; i++) {
            ans = min(ans, dp[i][n]);
            //cout<<"ans="<<ans<<endl;
        }

        return ans;
    }

    void output(int i, int j, int tmp) {
        cout<<"i="<<i<<" j="<<j<<" tmp="<<tmp<<endl;
    }

    void output2(int i, int j, int tmp) {
        cout<<"output2 i="<<i<<" j="<<j<<" tmp="<<tmp<<endl;
    }
};

第二种解法： 一维dp

思路: dp[n] = 1 + min{dp[n - i * i]} n - i * i >= 0

想使和为n，则最后一步使用的平方数可以为 1, 或者 4 或者 9, 即最后一步可以使用 i * i

则把所有可能枚举出来，求1 + min{dp[n - i * i]}

java代码

class Solution {
    //dp[i] = min{dp[i - j * j]} i - j * j >= 0
    public int numSquares(int n) {
        List<Integer> dp = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            dp.add(Integer.MAX_VALUE);
        }
        dp.set(0, 0);

        for (int i = 1; i <= n; i++) {
            int j = 1;
            while (i - j * j >= 0) {
                if (dp.get(i - j * j) != Integer.MAX_VALUE) {
                    dp.set(i, Math.min(dp.get(i), 1 + dp.get(i - j * j)));
                }
                ++j;
            }
        }

        return dp.get(n);
    }
}

c++代码

class Solution {
public:
    //dp[n] = min{dp[n - i*i]} i*i<=n
    //dp[n] represent least number of perfect square numbers
    //dp[0] = 0, dp[1] = 1, dp[2] = dp[1] + 1 = 2
    int numSquares(int n) {
        int dp[1 + n];
        for (int i = 0; i < n + 1; i++) {
            dp[i] = INT_MAX;
        }

        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            int j = 1;
            while (i - j * j >= 0) {
                if (dp[i - j * j] != INT_MAX) {
                    //cout<<dp[i]<<" "<<dp[i - j * j]<<endl;
                    dp[i] = min(dp[i], 1 + dp[i - j * j]);
                    //cout<<"dp[i]="<<dp[i]<<endl;
                }
                ++j;
            }
        }

        /*
        for (int d : dp) {
            cout<<d<<endl;
        }
        */

        return dp[n];
    }
};