cx_cs的博客

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要求：数组里的数拼接思路：跟剑指45题类似class Solution {public: string largestNumber(vector<int>& nums) { vector<string> ans; for(auto &num : nums) ans.push_back(to_string(num)); sort(ans.begin(), ans.end(), [](

要求：A代表1，给一个字符串思路：class Solution {public: int titleToNumber(string columnTitle) { int sum=0; int n=columnTitle.length(); for(int i=0;i<n;++i){ sum=sum*26+(columnTitle[i]-'A'+1); } return sum;

要求：包含其他字符思路：class Solution {public: bool isPalindrome(string s) { int n = s.size(); int left = 0, right = n - 1; while (left < right) { while (left < right && !isalnum(s[left])) { ++l

要求：对上一个字符串，读取连续个数思路：模拟或者打表class Solution {public: string countAndSay(int n) { string last="1"; for(int k=1;k<n;++k){ string cur; int len=last.length(); for(int i=0;i<len;){ c

要求：kmp算法，服了思路：复制粘贴https://leetcode-cn.com/problems/implement-strstr/solution/shi-xian-strstr-by-leetcode-solution-ds6y/class Solution {public: int strStr(string haystack, string needle) { int n = haystack.size(), m = needle.size();

要求：找起始下标思路：滑动窗口，击败90%class Solution {public: vector<int> findAnagrams(string s, string p) { vector<int> vp(26); for(char &c:p) vp[c-'a']++; int n=s.size(); int left=0,right=0; vecto

要求：输入：s = “3[a2[c]]”，输出：“accaccacc”思路：本题分两种情况，遇到字母或者数字[字母，注意考虑是否在嵌套下。流水线式代码。。时间100%class Solution {public: //3[a2[c2[d]e]]，栈不空的话弹出来要加栈顶 string decodeString(string s) { int n=s.length(); stack<int> si; stack<strin

要求：rt思路：滑动窗口，见注释class Solution {public: string minWindow(string s, string t) { unordered_map<char, int> hs, ht; for (auto c: t) ht[c] ++ ; string res; int cnt = 0; for (int i = 0, j = 0; i < s.size();

要求：生成匹配的括号序列思路：回溯class Solution {public: vector<string> ans; void dfs(string s,int left,int right,int& n){ if(s.size()==(n<<1)){ ans.push_back(s); return; } if(left<n)dfs(s+'(',l

要求：返回串思路：中心扩展class Solution {public: pair<int,int> expandAroundCenter(const string& s,int left,int right){ while(left>=0&&right<s.size()&&s[left]==s[right]){ --left; ++right; }

要求：rt思路：比较越界用214748364class Solution {public: int strToInt(string str) { int idx=0,n=str.length(); while(idx<n&&str[idx]==' ')++idx; int sign=1; if(idx<n&&str[idx]=='+'){ idx++;

要求：rt思路：class Solution {public: string reverseWords(string s) { string ans; int end=0; for(int i=s.length()-1;i>=0;){ while(i>=0&&s[i]==' ')--i; end=i; while(i>=0&&

要求：rt思路：字符串合并用a+b<b+aclass Solution {public: string minNumber(vector<int>& nums) { vector<string> ans; for(auto &num : nums) ans.push_back(to_string(num)); sort(ans.begin(), ans.end(), [](c

要求：rt思路：重复的先排序后last记录class Solution {public: vector<string> ans; string tmp; vector<bool> visited; void dfs(string& s){ if(tmp.length()==s.length()){ ans.push_back(tmp); return; }

要求：太复杂了思路：复制剑指官方题解class Solution {private: bool scanInteger(const string s, int& index){ if(s[index] == '+' || s[index] == '-') ++index; return scanUnsignedInteger(s, index);//跳过符号 } bool scanUnsignedInteger(c

要求：替换string空格思路：resize一下后插class Solution {public: string replaceSpace(string s) { int cnt=0; int n=s.length(); for(int i=0;i<n;++i){ if(s[i]==' ')++cnt; } s.resize(n+2*cnt); int j=n+2*cn

要求：挺简单的思路：class Solution {public: int countBinarySubstrings(string s) { int cnt=0; for(int i=0;i<s.size();){ char pre=s[i]; int precnt=0,curcnt=0; while(i<s.size()&&s[i]==pre){

要求：数思路：回文常用方法是拼接或者翻转class Solution {public: bool isPalindrome(int x) { if(x<0)return false; int xx=x; int sum=0; while(xx>=10){ sum=sum*10+xx%10; xx/=10; } return sum==x/1

要求：判断回文子串个数思路：动规，或者另一种中心扩展，从1个和2个为中心class Solution {public: int countSubstrings(string s) { int n=s.size(); vector<vector<int>> dp(n,vector<int>(n)); int num=n; for(int i=0;i<n-1;++i){

要求：s和t双射思路：class Solution {public: bool isIsomorphic(string s, string t) { if(s.size()!=t.size())return false; unordered_map<char,char> m; unordered_map<char,char> m2; for(int i=0;i<s.size();++i){

要求：能打乱顺序组合的最长回文思路：统计个数即可，但是，奇数次的可以不用完class Solution {public: int longestPalindrome(string s) { if(s=="")return 0; int len=0,oddnum=0; int n=s.size(); unordered_map<char,int> m; for(char c:s)

要求：每个字符出现个数是否相同，字符是unicode呢思路：unicode不写，c++难搞class Solution {public: bool isAnagram(string s, string t) { if(s.size()!=t.size())return false; int hash[26]={0}; for(char c:s) hash[c-'a']++; for(char c:t){