动态规划5步法初阶

class Solution {
public:
    int fib(int n) {
        if(n == 0)return 0;
        //确定动归数组
        vectordp(n+1);
        //确定递推公式
        //dp[i] = dp[i-1]+dp[i-2];
        //dp数组初始化
        dp[0] = 0;
        dp[1] = 1;
        //确定遍历顺序
        for(int i = 2;i<= n;i++){
            dp[i] = dp[i-1]+dp[i-2];
            cout << dp[i] << endl;
        }
        //举例推导dp数组
        //n = 10 ：0 1 1 2 3 5 8 13 21 34 55
        return dp[n];
    }
};

class Solution {
public:
    int climbStairs(int n) {
        if(n == 1)return 1;
        //确定动归数组
        vectordp(n+1);
        //确定递推公式
        //dp[i] = dp[i-1]+dp[i-2];
        //dp数组初始化
        dp[1] = 1;
        dp[2] = 2;
        //确定遍历顺序
        for(int i = 3;i<= n;i++){
            dp[i] = dp[i-1]+dp[i-2];
            cout << dp[i] << endl;
        }
        //举例推导dp数组
        //n = 10 ：0 1 1 2 3 5 8 13 21 34 55
        return dp[n];
    }
};

class Solution {
public:
    int minCostClimbingStairs(vector& cost) {
        int n = cost.size();
        if(n <= 1)return 0;
        //确定动归数组
        vectordp(n+1);
        //确定递推公式
        //dp[i] = min(dp[i-1]+cost[i],dp[i-2]+cost[i]);
        //dp数组初始化
        dp[0] = 0;
        dp[1] = 0;
        //确定遍历顺序
        for(int i = 2;i<= n;i++){
           dp[i] = min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2]);
           cout< 
不同路径I 
class Solution {
public:
    int uniquePaths(int m, int n) {
        if(n <= 1 || m <= 1)return min(m,n);
        //确定动归数组
        //vector>dp(m,vector(n,0));
        vector> dp(m, vector(n));
        //确定递推公式
        //dp[i,j] = dp[i][j-1] + dp[i-1][j];
        //dp数组初始化
        for(int i = 0;i< m;i++){
           dp[i][0] = 1;
        }
        for(int j = 0;j< n;j++){
           dp[0][j] = 1;
        }
        //确定遍历顺序
        for(int i = 1;i < m;i++){
            for(int j = 1;j < n;j++){
                dp[i][j] = dp[i][j-1] + dp[i-1][j];
                cout << dp[i][j] <<"    ";
            }
            cout << endl;
            
        }
        //举例推导dp数组
        //input:m = 3 n =3
        //output:
        
        return dp[m-1][n-1];
    }
};
 
不同路径II 
class Solution {
public:
    int uniquePathsWithObstacles(vector>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n =obstacleGrid[0].size();
        //确定动归数组
        vector> dp(m, vector(n,0));
        //确定递推公式
        //dp[i,j] = dp[i][j-1] + dp[i-1][j];
        //dp数组初始化
        for(int i = 0;i< m && obstacleGrid[i][0] != 1;i++){
             dp[i][0] = 1;
        }
        for(int j = 0;j< n &&obstacleGrid[0][j] != 1;j++){
           dp[0][j] = 1; 
        }
        //确定遍历顺序
        for(int i = 1;i < m;i++){
            for(int j = 1;j < n;j++){
                if(obstacleGrid[i][j] == 1){
                    dp[i][j] = 0;
                }
                else{
                    dp[i][j] = dp[i][j-1] + dp[i-1][j];
                }
                cout << dp[i][j] <<"    ";
            }
            cout << endl;
            
        }
        //举例推导dp数组
        //input: [[1,0,0],[0,1,0],[1,0,1],[1,0,0]]
        
        // output : 0
        
        return dp[m-1][n-1];
    }
};
 
整数拆分 
class Solution {
public:
    int integerBreak(int n) {
        if(n <= 1 )return 0;
        //确定动归数组
        vectordp(n+1);//从0到n，n+1个数都有属于自己的最大乘积
        
        //dp数组初始化
        //2 <= n <= 58，所以初始化可以不考虑0，1的情况
        dp[2] = 1;
        //确定遍历顺序
        //i>= 3 && i - j >= 2
        for(int i = 3;i <= n; i++){
            for(int j = 1;j <= i-2 ; j++){
                dp[i] = max(dp[i],max(j * dp[i-j],j * (i - j)));//这个不止是先找这两个最大，再和之前的比一下
                cout << dp[i] << "  ";
            }
            cout< 
不同二叉搜索树种数 
class Solution {
public:
    int numTrees(int n) {
        //确定动归数组
        vectordp(n+1);//从0到n，n+1个数都有对应的二叉搜索树种树
        //确定递推公式
        
        //dp数组初始化
        dp[0] = 1;
        //确定遍历顺序
        for(int i = 1;i <= n; i++){
            cout<< i <<": ";
            for (int j = 1; j <= i; j++) {
                dp[i] += dp[j - 1] * dp[i - j];
                cout<