LeetCode Brushing Notes-Dynamic Programming-day3

LeetCode Brushing Notes-Dynamic Programming-day3

198. Robbery

1. Topic

Link to the original title: 198. Robbery

2. Problem-solving ideas

We can use it f[i]to represent the total amount that the thief went to the first iroom to steal, which can be divided into two situations:

1. The thief does not choose ithe cash in the first room, thenf[i]=f[i-1];
2. The thief chooses to steal ithe cash in the first room, because consecutive rooms cannot be selected, sof[i]=f[i-2]+nums[i];

As shown in the picture:

Figure reproduced: https://www.acwing.com/solution/content/19504/

3. Code

class Solution {
    
    
public:
    int rob(vector<int>& nums) {
    
    
        int n=nums.size();
        vector<int> f(n+1);
        if(n<2) return nums[0];
        f[0]=nums[0],f[1]=max(nums[0],nums[1]);
        for(int i=2;i<n;i++){
    
    
            f[i]=max(f[i-1],f[i-2]+nums[i]);
        }
        return f[n-1];
    }
};

213. Dajia Palace II

1. Topic

Link to the original title: 213. House Robbery II

2. Problem-solving ideas

From the meaning of the question, we can see that this question has only one more condition compared to 198. Robbery : the first room and the last room cannot be selected at the same time , so we can enumerate these two situations:

1. Be sure not to choose the first one, that is, find [2,n]the maximum value that can be obtained in the interval
2. Be sure not to choose the last one, that is, find [1,n-1]the maximum value that can be obtained in the interval

Finally, take the maximum value of the two cases.

3. Code

class Solution {
    
    
public:
    int rob(vector<int>& nums) {
    
    
        int n=nums.size();
        vector<int> f(n+1);
        if(n<2) return nums[0];
        if(n==2) return max(nums[0],nums[1]);
        f[0]=nums[0],f[1]=max(nums[0],nums[1]);
        for(int i=2;i<n-1;i++){
    
    
            f[i]=max(f[i-1],f[i-2]+nums[i]);
        }
        int t1=f[n-2];
        f.clear();
        f[0]=0,f[1]=nums[1];
        for(int i=2;i<n;i++){
    
    
            f[i]=max(f[i-1],f[i-2]+nums[i]);
        }
        int t2=f[n-1];
        return max(t1,t2);
    }
};

740. Delete and get points

1. Topic

Original title link: 740. Delete and get points

2. Problem-solving ideas

Careful analysis of the title shows that this title is similar to 198. House robbery , the difference is that each element may have more than one.

Because the title requires that if you choose num[i], you cannot choose elements equal to nums[i] - 1and , which is equivalent to 198. We cannot choose consecutive rooms in house robbery.nums[i] + 1

We can use to maprecord the number of each element, and we can enumerate all elements from 1 to 10000 in the range of numbers given by the title. Use f[i]to represent the maximum number of points obtained by walking to this point, then:

f[i]=max(f[i-1],f[i-2]+i*cnt[i]);

3. Code

class Solution {
    
    
public:
    int deleteAndEarn(vector<int>& a) {
    
    
        map<int,int> cnt;
        for(auto x:a) cnt[x]++;
        vector<int> f(10001);
        f[1]=1*cnt[1];
        for(int i=2;i<10001;i++){
    
    
            f[i]=max(f[i-1],f[i-2]+i*cnt[i]);
        }
        return f[10000];
    }
};