【数据结构与算法】模拟

news2025/5/29 8:57:06

成熟不是为了走向复杂，而是为了抵达天真；不是为了变得深沉，而是为了保持清醒。

前言

这是我自己刷算法题的第五篇博客总结。

上一期笔记是关于前缀和算法：

【数据结构与算法】前缀和-CSDN博客https://blog.csdn.net/hsy1603914691/article/details/147953256?sharetype=blogdetail&sharerId=147953256&sharerefer=PC&sharesource=hsy1603914691&spm=1011.2480.3001.8118

技巧

1. 模拟题思路比较简单，先模拟算法流程，再把流程转换成代码。

2. 模拟题需要特别注意开始和结束，可能需要特殊处理。

3. 模拟题的重点是找规律。

例题

1. leetcode-724题：

1576. 替换所有的问号 - 力扣（LeetCode）https://leetcode.cn/problems/replace-all-s-to-avoid-consecutive-repeating-characters/description/

class Solution {
public:
    string modifyString(string s) 
    {
        int cur=0;
        while(cur<s.size())
        {
            if(s[cur]=='?')
            {
                for(char replace='a';replace<='z';replace++)
                {
                    if(cur==0&&replace!=s[cur+1])
                        s[cur]=replace;
                    else if(cur==s.size()-1&&replace!=s[cur-1])
                        s[cur]=replace;
                    else if(cur!=0&&cur!=s.size()-1&&replace!=s[cur+1]&&replace!=s[cur-1])
                        s[cur]=replace;
                    else
                    {}
                }
            }
            cur++;
        }
         return s;
    }
};

2. leetcode-495题：

495. 提莫攻击 - 力扣（LeetCode）https://leetcode.cn/problems/teemo-attacking/description/

class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) 
    {
        int cur=0,sum=0;
        while(cur<timeSeries.size())
        {
            if((cur!=timeSeries.size()-1&&timeSeries[cur]+duration<=timeSeries[cur+1])||(cur==timeSeries.size()-1))
            {
                sum+=duration;
                cur++;
            }
            else
            {
                sum+=timeSeries[cur+1]-timeSeries[cur];
                cur++;
            }
        }
        return sum;
    }
};

3. leetcode-6题：

6. Z 字形变换 - 力扣（LeetCode）https://leetcode.cn/problems/zigzag-conversion/

class Solution {
public:
    string convert(string s, int numRows) 
    {
        if(numRows==1)
            return s;
        string ret;
        int d=2*numRows-2,n=s.size();
        for(int i=0;i<n;i+=d)
            ret+=s[i];
        for(int k=1;k<numRows-1;k++)
        {
            for(int i=k,j=d-k;i<n||j<n;i+=d,j+=d)
            {
                if(i<n)
                    ret+=s[i];
                if(j<n)
                    ret+=s[j];
            }
        }
        for(int i=numRows-1;i<n;i+=d)
            ret+=s[i];
        return ret;
    }
};

4. leetcode-38题：

38. 外观数列 - 力扣（LeetCode）https://leetcode.cn/problems/count-and-say/

class Solution {
public:
    string countAndSay(int n) 
    {
        string s="1";
        for(int i=1;i<n;i++)
        {
            string temp;
            int left=0,right=0,count=0;
            while(right<s.size())
            {
                while(s[right]==s[left]&&right<s.size())
                    right++;
                count=right-left;
                temp.push_back(count+'0');
                temp.push_back(s[left]);
                left=right;
            }
            s=temp;
        }
        return s;
    }
};

5. leetcode-1419题：

1419. 数青蛙 - 力扣（LeetCode）https://leetcode.cn/problems/minimum-number-of-frogs-croaking/

class Solution {
public:
    int minNumberOfFrogs(string croakOfFrogs) 
    {
        unordered_map<char,int> m;
        int right=0;
        while(right<croakOfFrogs.size())
        {
            if(croakOfFrogs[right]=='c')
            {
                if(m['k']==0)
                    m['c']++;
                else
                    m['k']--,m['c']++;
            }
            else if(croakOfFrogs[right]=='r')
            {
                if(m['c']==0)
                    return -1;
                else
                    m['c']--,m['r']++;
            }
            else if(croakOfFrogs[right]=='o')
            {
                if(m['r']==0)
                    return -1;
                else
                    m['r']--, m['o']++;
            }
            else if(croakOfFrogs[right]=='a')
            {
                if(m['o']==0)
                    return -1;
                else
                    m['o']--,m['a']++;
            }
            else if(croakOfFrogs[right]=='k')
            {
                if(m['a']==0)
                    return -1;
                else
                    m['a']--,m['k']++;
            }
            else
                return -1;

            right++;
        }
        if(m['c']==0&&m['r']==0&&m['o']==0&&m['a']==0)
            return m['k'];
        else
            return -1;
    }
};