leetcode Integer Break

 

 

 

动态规划问题解题思路： 

 1.递归解法（效率偏低）

/// 343. Integer Break
/// https://leetcode.com/problems/integer-break/description/
/// 暴力搜索
/// 在Leetcode中提交这个版本的代码会超时! (Time Limit Exceeded)
/// 时间复杂度: O(n^n)
/// 空间复杂度: O(n)
public class Solution1 {

    public int integerBreak(int n) {

        if(n < 1)
            throw new IllegalArgumentException("n should be greater than zero");

        return breakInteger(n);
    }

    // 将n进行分割(至少分割两部分), 可以获得的最大乘积
    private int breakInteger(int n){

        if(n == 1)
            return 1;

        int res = -1;
        for(int i = 1 ; i <= n - 1 ; i ++)
            res = max3(res, i * (n - i), i * breakInteger(n - i));
        return res;
    }

    private int max3(int a, int b, int c){
        return Math.max(a, Math.max(b, c));
    }

    public static void main(String[] args) {

        System.out.println((new Solution1()).integerBreak(2));
        System.out.println((new Solution1()).integerBreak(10));
    }
}

2.记忆化搜索解法

import java.util.Arrays;

/// 343. Integer Break
/// https://leetcode.com/problems/integer-break/description/
/// 记忆化搜索
/// 时间复杂度: O(n^2)
/// 空间复杂度: O(n)
public class Solution2 {

    private int[] memo;

    public int integerBreak(int n) {

        if(n < 1)
            throw new IllegalArgumentException("n should be greater than zero");

        memo = new int[n+1];
        Arrays.fill(memo, -1);

        return breakInteger(n);
    }

    // 将n进行分割(至少分割两部分), 可以获得的最大乘积
    private int breakInteger(int n){

        if(n == 1)
            return 1;

        if(memo[n] != -1)
            return memo[n];

        int res = -1;
        for(int i = 1 ; i <= n - 1 ; i ++)
            res = max3(res, i * (n - i) , i * breakInteger(n - i));
        memo[n] = res;
        return res;
    }

    private int max3(int a, int b, int c){
        return Math.max(a, Math.max(b, c));
    }

    public static void main(String[] args) {

        System.out.println((new Solution2()).integerBreak(2));
        System.out.println((new Solution2()).integerBreak(10));
    }

3.动态规划解法

/// 343. Integer Break
/// https://leetcode.com/problems/integer-break/description/
/// 动态规划
/// 时间复杂度: O(n^2)
/// 空间复杂度: O(n)
public class Solution3 {

    public int integerBreak(int n) {

        if(n < 1)
            throw new IllegalArgumentException("n should be greater than zero");

        int[] memo = new int[n+1];
        memo[1] = 1;
        for(int i = 2 ; i <= n ; i ++)
            // 求解memo[i]
            for(int j = 1 ; j <= i - 1 ; j ++)
                memo[i] = max3(memo[i], j * (i - j), j * memo[i - j]);

        return memo[n];
    }

    private int max3(int a, int b, int c){
        return Math.max(a, Math.max(b, c));
    }

    public static void main(String[] args) {

        System.out.println((new Solution3()).integerBreak(2));
        System.out.println((new Solution3()).integerBreak(10));
    }
}

1. 279 Perfect Squares

2. 91 Decode Ways

 3.Unique Paths

4. Unique Paths