leetcode Integer Break

偏执的太偏执、 2022-04-11 11:23 71阅读 0赞

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70][]

![20181205223800676.png][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 1][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 2][]

动态规划问题解题思路：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 3][]

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

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 4][]

2. 91 Decode Ways

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 5][]

3.Unique Paths

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 6][]

4. Unique Paths

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 7][]

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70]: /images/20220411/e3576d0d230c4de48740f7f316241ab4.png
[20181205223800676.png]: /images/20220411/a0d22c7b897349d6b986f7226b5e4c66.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 1]: /images/20220411/8c92501437f14ba9af42d08eca8fed40.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 2]: /images/20220411/b173e113a7da4506aed2dd19acf71a71.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 3]: /images/20220411/4fb0bb9f9ac3434ba2631218039638a1.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 4]: /images/20220411/4f9ffc3699844c97af9014ad6e1b969c.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 5]: /images/20220411/b441322cf41944f493bfaadf37d41212.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 6]: /images/20220411/7cffcedde26e46bbb4b3e88089fb2aa2.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1Mzk0ODkx_size_16_color_FFFFFF_t_70 7]: /images/20220411/3d79291fd2b746548a68a1faac21c074.png