力扣 C++|一题多解之动态规划专题(1)

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动态规划

Dynamic Programming

简写为 DP,是运筹学的一个分支,是求解决策过程最优化的过程。20世纪50年代初,美国数学家贝尔曼(R.Bellman)等人在研究多阶段决策过程的优化问题时,提出了著名的最优化原理,从而创立了动态规划。动态规划的应用极其广泛,包括工程技术、经济、工业生产、军事以及自动化控制等领域,并在背包问题、生产经营问题、资金管理问题、资源分配问题、最短路径问题和复杂系统可靠性问题等中取得了显著的效果。

动态规划算法的基本步骤包括:

  1. 确定状态:确定需要求解的状态,并将其表示为变量。
  2. 确定状态转移方程:根据问题的特定约束条件和目标函数,确定状态之间的转移关系,并将其表示为数学公式。
  3. 初始化:为初始状态赋初值,并将其表示为初始条件。
  4. 递推计算:根据状态转移方程,使用循环依次计算各个状态的解,并将其保存在数组或表中。
  5. 求解最终结果:根据问题的目标,从计算得到的解中得出最终结果。

动态规划算法可以用于解决各种问题,例如最短路径问题、背包问题、最长公共子序列问题等。在实现动态规划算法时,需要根据具体问题的特点进行设计和调整,以确保算法的正确性和效率。

适用条件

任何思想方法都有一定的局限性,超出了特定条件,它就失去了作用。同样,动态规划也并不是万能的。适用动态规划的问题必须满足最优化原理和无后效性。

最优化原理(最优子结构性质)

最优化原理可这样阐述:一个最优化策略具有这样的性质,不论过去状态和决策如何,对前面的决策所形成的状态而言,余下的诸决策必须构成最优策略。简而言之,一个最优化策略的子策略总是最优的。一个问题满足最优化原理又称其具有最优子结构性质 [8] 。

无后效性

将各阶段按照一定的次序排列好之后,对于某个给定的阶段状态,它以前各阶段的状态无法直接影响它未来的决策,而只能通过当前的这个状态。换句话说,每个状态都是过去历史的一个完整总结。这就是无后向性,又称为无后效性 [8] 。

子问题的重叠性

动态规划算法的关键在于解决冗余,这是动态规划算法的根本目的。动态规划实质上是一种以空间换时间的技术,它在实现的过程中,不得不存储产生过程中的各种状态,所以它的空间复杂度要大于其他的算法。选择动态规划算法是因为动态规划算法在空间上可以承受,而搜索算法在时间上却无法承受,所以我们舍空间而取时间。

真题举例(1)

44. 通配符匹配

给定一个字符串 (s) 和一个字符模式 (p) ,实现一个支持 '?' 和 '*' 的通配符匹配。
'?' 可以匹配任何单个字符。'*' 可以匹配任意字符串(包括空字符串)。
两个字符串完全匹配才算匹配成功。

说明:

  • s 可能为空,且只包含从 a-z 的小写字母。
  • p 可能为空,且只包含从 a-z 的小写字母,以及字符 ? 和 *

示例 1:

输入:s = "aa"  p = "a"
输出: false
解释: "a" 无法匹配 "aa" 整个字符串。

示例 2:

输入:s = "aa"  p = "*"
输出: true
解释: '*' 可以匹配任意字符串。

示例 3:

输入:s = "cb"  p = "?a"
输出: false
解释: '?' 可以匹配 'c', 但第二个 'a' 无法匹配 'b'。

示例 4:

输入:s = "adceb"  p = "*a*b"
输出: true
解释: 第一个 '*' 可以匹配空字符串, 第二个 '*' 可以匹配字符串 "dce".

示例 5:

输入:s = "acdcb"  p = "a*c?b"
输出: false

代码1:

#include 
using namespace std;

class Solution
{
public:
    bool isMatch(string s, string p)
    {
        vector> dp(s.size() + 1, vector(p.size() + 1));
        dp[0][0] = 1;
        for (int j = 1; j <= p.size(); j++)
        {
            dp[0][j] = dp[0][j - 1] && p[j - 1] == '*';
        }
        for (int i = 1; i <= s.size(); i++)
        {
            for (int j = 1; j <= p.size(); j++)
            {
                if (p[j - 1] == '*')
                {
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                }
                else
                {
                    dp[i][j] = (s[i - 1] == p[j - 1] || p[j - 1] == '?') && dp[i - 1][j - 1];
                }
            }
        }
        return dp[s.size()][p.size()];
    }
};

int main()
{
	Solution s;

	cout << s.isMatch("aa", "a") << endl;
	cout << s.isMatch("aa", "*") << endl;
	cout << s.isMatch("cb", "?a") << endl;
	cout << s.isMatch("adceb", "*a*b") << endl;
	cout << s.isMatch("acdcb", "a*c?b") << endl;
	
	return 0;
} 

代码2: 

#include 
using namespace std;

class Solution
{
public:
    bool isMatch(string s, string p)
    {
        int m = s.size();
        int n = p.size();

        vector> dp(m + 1, vector(n + 1));
        dp[0][0] = true;

        for (int i = 1; i <= n; i++)
        {
            if (p[i - 1] == '*')
                dp[0][i] = true;
            else
                break;
        }

        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {

                if (p[j - 1] == '*')
                {
                    dp[i][j] |= dp[i][j - 1];
                    dp[i][j] |= dp[i - 1][j];
                }
                else
                {
                    if (p[j - 1] == '?' || s[i - 1] == p[j - 1])
                    {
                        dp[i][j] |= dp[i - 1][j - 1];
                    }
                }
            }
        }

        return dp[m][n];
    }
};

int main()
{
	Solution s;

	cout << s.isMatch("aa", "a") << endl;
	cout << s.isMatch("aa", "*") << endl;
	cout << s.isMatch("cb", "?a") << endl;
	cout << s.isMatch("adceb", "*a*b") << endl;
	cout << s.isMatch("acdcb", "a*c?b") << endl;
	
	return 0;
} 

代码3: 

#include 
using namespace std;

class Solution
{
public:
    bool isMatch(string s, string p)
    {
        if (p.empty())
            return s.empty();
        if (s.empty())
        {
            if (p[0] == '*')
                return isMatch(s, p.substr(1));
            else
                return false;
        }

        if (p[0] == '*')
            return isMatch(s, p.substr(1)) || isMatch(s.substr(1), p);
        else
            return (s[0] == p[0] || p[0] == '?') && isMatch(s.substr(1), p.substr(1));
    }
};

int main()
{
	Solution s;

	cout << s.isMatch("aa", "a") << endl;
	cout << s.isMatch("aa", "*") << endl;
	cout << s.isMatch("cb", "?a") << endl;
	cout << s.isMatch("adceb", "*a*b") << endl;
	cout << s.isMatch("acdcb", "a*c?b") << endl;
	
	return 0;
} 

代码4: 

#include 
using namespace std;

class Solution {
public:
    bool isMatch(string s, string p) {
        if (s == "" && p == "" || (s == "" && p == "*"))
            return true;
        if (s == p)
            return true;
        int lens = s.length();
        int lenp = p.length();
        bool questionm = false, starm = false;
        for (int k = 0; k < lenp; k++) {
            if (p[k] == '?')
                questionm = true;
            if (p[k] == '*')
                starm = true;
        }
        if (lenp != lens && questionm == false && starm == false)
            return false;
        int i = 0, j = 0;
        int mstar = 0, sstar = -1;
        while (i < lens) {
            if (j < lenp && p[j] == '*') {
                mstar = i;
                sstar = j;
                j += 1;
            } else if (j < lenp && (s[i] == p[j] || p[j] == '?')) {
                i++;
                j++;
            } else if (sstar != -1) {
                mstar += 1;
                j = sstar + 1;
                i = mstar;
            } else
                return false;
        }
        while (j < lenp) {
            if (p[j] != '*')
                return false;
            j++;
        }
        return true;
    }
};

int main()
{
	Solution s;

	cout << s.isMatch("aa", "a") << endl;
	cout << s.isMatch("aa", "*") << endl;
	cout << s.isMatch("cb", "?a") << endl;
	cout << s.isMatch("adceb", "*a*b") << endl;
	cout << s.isMatch("acdcb", "a*c?b") << endl;
	
	return 0;
} 

62. 不同路径

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为 “Start” )。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为 “Finish” )。问总共有多少条不同的路径?

示例 1:

力扣 C++|一题多解之动态规划专题(1)_第1张图片

输入:m = 3, n = 7
输出:28

示例 2:

输入:m = 3, n = 2
输出:3
解释:从左上角开始,总共有 3 条路径可以到达右下角。
1. 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右
3. 向下 -> 向右 -> 向下

示例 3:

输入:m = 7, n = 3
输出:28

示例 4:

输入:m = 3, n = 3
输出:6

提示:

  • 1 <= m, n <= 100
  • 题目数据保证答案小于等于 2 * 10^9

代码1:

#include 
using namespace std;

class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        int N = m + n - 2;
        int M = m < n ? m - 1 : n - 1;

        long ans = 1;
        for (int i = 1; i <= M; i++)
            ans = ans * (N - i + 1) / i;
        return ans;
    }
};

int main()
{
	Solution s;
	cout << s.uniquePaths(3, 7) << endl;
	cout << s.uniquePaths(3, 2) << endl;
	cout << s.uniquePaths(7, 3) << endl;
	cout << s.uniquePaths(3, 3) << endl;
	
	return 0;
} 

代码2: 

#include 
using namespace std;

class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        if (m <= 0 || n <= 0)
        {
            return 0;
        }

        vector> dp(m + 1, vector(n + 1, 0));
        for (int i = 0; i < m; i++)
        {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++)
        {
            dp[0][i] = 1;
        }

        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	cout << s.uniquePaths(3, 7) << endl;
	cout << s.uniquePaths(3, 2) << endl;
	cout << s.uniquePaths(7, 3) << endl;
	cout << s.uniquePaths(3, 3) << endl;
	
	return 0;
} 

代码3: 

#include 
using namespace std;

typedef vector BigInt;
class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        if (m == 0 || n == 0)
            return 0;
        if (m == 1 || n == 1)
            return 1;
        int m_ = m - 1 + n - 1;
        int n_ = n - 1;
        BigInt a = fac(m_);
        int result = 0;
        for (int i = n_; i >= 1; i--)
            a = div(a, i);
        for (int i = m_ - n_; i >= 1; i--)
            a = div(a, i);
        int k = a.size() - 1;
        while (a[k] == 0)
            k--;
        for (int i = k; i >= 0; i--)
            result = result * 10 + a[i];

        return result;
    }

    BigInt fac(int n)
    {
        BigInt result;
        result.push_back(1);
        for (int factor = 1; factor <= n; ++factor)
        {
            long long carry = 0;
            for (auto &item : result)
            {
                long long product = item * factor + carry;
                item = product % 10;
                carry = product / 10;
            }
            if (carry > 0)
            {
                while (carry > 0)
                {
                    result.push_back(carry % 10);
                    carry /= 10;
                }
            }
        }
        return result;
    }
    BigInt div(BigInt a, int d)
    {
        int b = 0;
        BigInt result;
        int len = a.size();
        for (int i = len - 1; i >= 0; i--)
        {
            b = b * 10 + a[i];
            result.insert(result.begin(), b / d);
            b = b % d;
        }
        return result;
    }
};

int main()
{
	Solution s;
	cout << s.uniquePaths(3, 7) << endl;
	cout << s.uniquePaths(3, 2) << endl;
	cout << s.uniquePaths(7, 3) << endl;
	cout << s.uniquePaths(3, 3) << endl;
	
	return 0;
} 

代码4: 

#include 
using namespace std;

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector> path(m, vector(n, 0));
        for (int i = 0; i < n; i++)
            path[0][i] = 1;
        for (int i = 0; i < m; i++)
            path[i][0] = 1;
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                path[i][j] = path[i - 1][j] + path[i][j - 1];
        return path[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	cout << s.uniquePaths(3, 7) << endl;
	cout << s.uniquePaths(3, 2) << endl;
	cout << s.uniquePaths(7, 3) << endl;
	cout << s.uniquePaths(3, 3) << endl;
	
	return 0;
} 

63. 不同路径 II

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” )。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为“Finish”)。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径?

力扣 C++|一题多解之动态规划专题(1)_第2张图片

网格中的障碍物和空位置分别用 1 和 0 来表示。

示例 1:

力扣 C++|一题多解之动态规划专题(1)_第3张图片

输入:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
输出:2
解释:3x3 网格的正中间有一个障碍物。从左上角到右下角一共有 2 条不同的路径:
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

示例 2:

力扣 C++|一题多解之动态规划专题(1)_第4张图片

输入:obstacleGrid = [[0,1],[0,0]]
输出:1

提示:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] 为 0 或 1

代码1: 

#include 
using namespace std;

class Solution
{
public:
    int uniquePathsWithObstacles(vector> &obstacleGrid)
    {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int p[m][n];

        int k = 0;
        while (k < m && obstacleGrid[k][0] != 1)
            p[k++][0] = 1;

        while (k < m)
            p[k++][0] = 0;

        k = 0;
        while (k < n && obstacleGrid[0][k] != 1)
            p[0][k++] = 1;
        while (k < n)
            p[0][k++] = 0;

        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
            {
                if (obstacleGrid[i][j] == 1)
                    p[i][j] = 0;
                else
                    p[i][j] = p[i - 1][j] + p[i][j - 1];
            }
        return p[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	vector> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;

	obstacleGrid = {{0,1},{0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
	
	return 0;
} 

代码2: 

#include 
using namespace std;

class Solution
{
public:
    int uniquePathsWithObstacles(vector> &obstacleGrid)
    {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector> dp(m, vector(n, 0));
        for (int i = 0; i < m && obstacleGrid[i][0] != 1; i++)
        {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n && obstacleGrid[0][i] != 1; i++)
        {
            dp[0][i] = 1;
        }
        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                if (obstacleGrid[i][j] != 1)
                {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	vector> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;

	obstacleGrid = {{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
	
	return 0;
} 

代码3: 

#include 
using namespace std;

class Solution
{
public:
    int uniquePathsWithObstacles(vector> &obstacleGrid)
    {
        if (obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
            return 0;
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector> info(m, vector(n, 0));
        for (int i = 0; i < m; ++i)
        {
            if (obstacleGrid[i][0] == 1)
            {
                for (int j = i; j < m; j++)
                {
                    info[j][0] = 0;
                }
                break;
            }
            else
                info[i][0] = 1;
        }
        for (int i = 0; i < n; ++i)
        {
            if (obstacleGrid[0][i] == 1)
            {
                for (int j = i; j < n; ++j)
                {
                    info[0][j] = 0;
                }
                break;
            }
            else
                info[0][i] = 1;
        }
        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                if (obstacleGrid[i][j] == 1)
                {
                    info[i][j] = 0;
                }
                else
                {
                    info[i][j] = info[i - 1][j] + info[i][j - 1];
                }
            }
        }
        return info[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	vector> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;

	obstacleGrid = {{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
	
	return 0;
} 

代码4: 

#include 
using namespace std;

class Solution {
public:
    int uniquePathsWithObstacles(vector> &obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        if (obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1)
            return 0;
        vector> dp(m, vector(n, 0));
        dp[0][0] = 1;
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 0)
                dp[i][0] = dp[i - 1][0];
        }
        for (int i = 1; i < n; i++) {
            if (obstacleGrid[0][i] == 0)
                dp[0][i] = dp[0][i - 1];
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 0)
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};

int main()
{
	Solution s;
	vector> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;

	obstacleGrid = {{0,1,0},{0,0,0}};
	cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
	
	return 0;
} 

64. 最小路径和

给定一个包含非负整数的 m x n 网格 grid ,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小。

说明:每次只能向下或者向右移动一步。

示例 1:

力扣 C++|一题多解之动态规划专题(1)_第5张图片

输入:grid = [[1,3,1],[1,5,1],[4,2,1]]
输出:7
解释:因为路径 1→3→1→1→1 的总和最小。

示例 2:

输入:grid = [[1,2,3],[4,5,6]]
输出:12

提示:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 200
  • 0 <= grid[i][j] <= 100

代码1:

#include 
using namespace std;

class Solution
{
public:
    int minPathSum(vector> &grid)
    {
        if (grid.size() == 0)
            return 0;
        int m = grid.size();
        int n = grid[0].size();
        vector> m_memo = vector>(m + 1, vector(n + 1, 0));

        for (int i = n - 1; i >= 0; --i)
            m_memo[m - 1][i] = grid[m - 1][i] + m_memo[m - 1][i + 1];
        for (int j = m - 1; j >= 0; --j)
            m_memo[j][n - 1] = grid[j][n - 1] + m_memo[j + 1][n - 1];

        for (int i = m - 2; i >= 0; --i)
        {
            for (int j = n - 2; j >= 0; --j)
            {
                m_memo[i][j] = grid[i][j] + min(m_memo[i][j + 1], m_memo[i + 1][j]);
            }
        }
        return m_memo[0][0];
    }
};

int main()
{
	Solution s;
	vector> grid = {{1,3,1},{1,5,1},{4,2,1}};
	cout << s.minPathSum(grid) << endl;
	
	grid = {{1,2,3},{4,5,6}};
	cout << s.minPathSum(grid) << endl;
	
	return 0;
} 

代码2: 

#include 
using namespace std;

class Solution
{
public:
    int minPathSum(vector> &grid)
    {
        int row = grid.size();
        int column = grid[0].size();
        for (int i = 1; i < column; ++i)
        {
            grid[0][i] = grid[0][i - 1] + grid[0][i];
        }
        for (int i = 1; i < row; ++i)
        {
            grid[i][0] = grid[i - 1][0] + grid[i][0];
        }
        for (int i = 1; i < row; ++i)
        {
            for (int j = 1; j < column; ++j)
            {
                int temp = grid[i - 1][j] > grid[i][j - 1] ? grid[i][j - 1] : grid[i - 1][j];
                grid[i][j] = grid[i][j] + temp;
            }
        }
        return grid[row - 1][column - 1];
    }
};

int main()
{
	Solution s;
	vector> grid = {{1,3,1},{1,5,1},{4,2,1}};
	cout << s.minPathSum(grid) << endl;
	
	grid = {{1,2,3},{4,5,6}};
	cout << s.minPathSum(grid) << endl;
	
	return 0;
} 

代码3: 

#include 
using namespace std;

class Solution
{
public:
    int minPathSum(vector> &grid)
    {
        int row = grid.size();
        int col = grid[0].size();
        vector f(col, 0);

        for (int i = 0; i < row; ++i)
        {
            f[0] = f[0] + grid[i][0];
            for (int j = 1; j < col; ++j)
            {
                if (i == 0)
                    f[j] = f[j - 1] + grid[i][j];
                else
                    f[j] = min(f[j - 1], f[j]) + grid[i][j];
            }
        }
        return f[col - 1];
    }
};

int main()
{
	Solution s;
	vector> grid = {{1,3,1},{1,5,1},{4,2,1}};
	cout << s.minPathSum(grid) << endl;
	
	grid = {{1,2,3},{4,5,6}};
	cout << s.minPathSum(grid) << endl;
	
	return 0;
} 

代码4: 

#include 
using namespace std;

class Solution {
private:
    int m, n;
    int memo[100][100];
public:
    int minPathSum(vector> &grid) {
        m = grid.size(), n = grid[0].size();
        for (int i = 0; i < m; i++) {
            memset(memo[i], -1, sizeof(int) * n);
        }
        return dfs(grid, 0, 0);
    }
    int dfs(vector> &grid, int r, int c) {
        if (r < 0 || r >= m || c < 0 || c >= n)
            return 1000000;
        if (memo[r][c] != -1)
            return memo[r][c];
        if (r == m - 1 && c == n - 1) {
            memo[r][c] = grid[m - 1][n - 1];
            return memo[r][c];
        }
        int right = dfs(grid, r, c + 1);
        int down = dfs(grid, r + 1, c);
        memo[r][c] = min(right, down) + grid[r][c];
        return memo[r][c];
    }
};

int main()
{
	Solution s;
	vector> grid = {{1,3,1},{1,5,1},{4,2,1}};
	cout << s.minPathSum(grid) << endl;
	
	grid = {{1,2,3},{4,5,6}};
	cout << s.minPathSum(grid) << endl;
	
	return 0;
} 

续:https://hannyang.blog.csdn.net/article/details/132091605

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