参考练习习题总集

滑动窗口太简单了，没啥说的自己做吧。

3. 无重复字符的最长子串

class Solution {
public:int lengthOfLongestSubstring(string s) {if (s.size()==0) return 0;unordered_map<char,int> zd;zd[s[0]]=0;int max_len=1;for (int i=1;i<s.size();i++)if (zd.find(s[i])==zd.end()) zd[s[i]]=i;else{int x=my_min(zd);if (i-x>max_len) max_len=i-x;dlt(zd,zd[s[i]]);zd[s[i]]=i;}if (zd.size()>max_len) max_len=zd.size();return max_len;}int my_min(unordered_map<char,int> & zd){int x=5*pow(10,4)+1;for (unordered_map<char,int>::iterator zz=zd.begin();zz!=zd.end();zz++)if (zz->second<x) x=zz->second;return x;}void dlt(unordered_map<char,int> & zd,int x){vector<char> lt;for (unordered_map<char,int>::iterator zz=zd.begin();zz!=zd.end();zz++)if (zz->second<x) lt.push_back(zz->first);for (vector<char>::iterator zz=lt.begin();zz!=lt.end();zz++)zd.erase(*zz);}
};

30. 串联所有单词的子串

class Solution {
public:int word_len;unordered_map<string,int> zd;vector<int> findSubstring(string s, vector<string>& words) {vector<int> result;int length=words.size()*words[0].size();if (s.size()<length) return result;unordered_map<char,int> zd1,zd2;for (int i=0;i<words.size();i++)for (int j=0;j<words[i].size();j++){if (zd1.find(words[i][j])==zd1.end()) zd1[words[i][j]]=0;zd1[words[i][j]]+=1;}for (int i=0;i<length;i++){if (zd2.find(s[i])==zd2.end()) zd2[s[i]]=0;zd2[s[i]]+=1;}word_len=words[0].size();for (int i=0;i<words.size();i++){if (zd.find(words[i])==zd.end()) zd[words[i]]=0;zd[words[i]]+=1; }string temp;temp=s.substr(0,length);if (zd1==zd2 and func(temp,zd)) result.push_back(0);for (int i=1;i<s.size()-length+1;i++){zd2[s[i-1]]-=1;if (zd2[s[i-1]]==0) zd2.erase(s[i-1]);if (zd2.find(s[i+length-1])==zd2.end()) zd2[s[i+length-1]]=0;zd2[s[i+length-1]]+=1;temp=s.substr(i,length);if (zd1==zd2 and func(temp,zd)) result.push_back(i);}return result;}bool func(const string & s,unordered_map<string,int> zd){for (int i=0;i<s.size();i+=word_len){string temp=s.substr(i,word_len);if (zd.find(temp)==zd.end()) return false;zd[temp]-=1;if (zd[temp]==0) zd.erase(temp);}return true;}
};

76. 最小覆盖子串

class Solution {
public:string minWindow(string s, string t) {string result;unordered_map<char,int> zd;for (int i=0;i<t.size();i++){if (zd.find(t[i])==zd.end()) zd[t[i]]=0;zd[t[i]]+=1;}vector<pair<char,int>> lb;for (int i=0;i<s.size();i++)if (zd.find(s[i])!=zd.end())lb.push_back({s[i],i});if (lb.size()<t.size()) return result;for (int i=0;i<t.size();i++)zd[lb[i].first]-=1;int l=0,r=t.size();while (!check(zd)){if (r>=lb.size()) return result;zd[lb[r].first]-=1;r+=1;}while (zd[lb[l].first]+1<=0){zd[lb[l].first]+=1;l+=1;}result=s.substr(lb[l].second,lb[r-1].second-lb[l].second+1);while (1){zd[lb[l].first]+=1;l+=1;while (zd[lb[l-1].first]>0){if (r>=lb.size()) return result;zd[lb[r].first]-=1;r+=1;}if (lb[r-1].second-lb[l].second+1<result.size())result=s.substr(lb[l].second,lb[r-1].second-lb[l].second+1);}return result;}bool check(const unordered_map<char,int> & zd){for (auto zz=zd.begin();zz!=zd.end();zz++)if (zz->second>0) return false;return true;}
};

187. 重复的DNA序列

class Solution {
public:vector<string> findRepeatedDnaSequences(string s) {vector<string> lb;if (s.size()<=10) return lb;unordered_map<string,int> zd;string s_new=s.substr(0,10);zd[s_new]=1;for (int i=10;i<s.size();i++){s_new.erase(0,1);s_new+=s[i];if (zd.find(s_new)==zd.end()) zd[s_new]=1;else{if (zd[s_new]==1) lb.push_back(s_new);zd[s_new]+=1;}}return lb;}
};

219. 存在重复元素 II

直接暴力

class Solution {
public:bool containsNearbyDuplicate(vector<int>& nums, int k) {for (int i=0;i<nums.size();i++)for (int j=i+1;j<=i+k and j<nums.size();j++)if (nums[i]==nums[j]) return 1;return 0;}
};

220. 存在重复元素 III

首次碰见这种思想，我就直接抄的题解
又学到了一种思想

class Solution {
public:int size;bool containsNearbyAlmostDuplicate(vector<int>& nums, int indexDiff, int valueDiff) {size=valueDiff+1;unordered_map<int,int> zd;for (int i=0;i<nums.size();i++) {int u=nums[i],idx=getIdx(u);if (zd.find(idx)!=zd.end()) return true;int l=idx-1,r=idx+1;if (zd.find(l)!=zd.end() and abs(u-zd[l])<=valueDiff) return true;if (zd.find(r)!=zd.end() and abs(u-zd[r])<=valueDiff) return true;zd[idx]=u;if (i>=indexDiff) zd.erase(getIdx(nums[i-indexDiff]));}return false;}int getIdx(int u) {return u>=0?u/size:(u+1)/size-1;}
};

396. 旋转函数

class Solution {
public:int maxRotateFunction(vector<int>& nums) {int count=0;for (int x:nums) count+=x;int f0=0;for (int i=0;i<nums.size();i++)f0+=i*nums[i];int result=f0;for (int i=1;i<nums.size();i++){f0=f0+count-nums[nums.size()-i]*nums.size();if (f0>result) result=f0;}return result;}
};

424. 替换后的最长重复字符

我好菜啊

class Solution {
public:int characterReplacement(string s, int k) {int l=0,r=0,result=-1,result_temp;unordered_map<char,int> zd;while (r!=s.size()){while (1){if (r-l<k+1) break;char temp=find_max_char(zd);if (r-l-zd[temp]<k) break;if (s[r]==temp) break;l+=1;zd[s[l-1]]-=1;if (zd[s[l-1]]==0) zd.erase(s[l-1]);}if (zd.find(s[r])==zd.end()) zd[s[r]]=0;zd[s[r]]+=1;r+=1;result_temp=r-l;if (result_temp>result) result=result_temp;}return result;}char find_max_char(unordered_map<char,int> & zd){char result=zd.begin()->first;for (auto zz=++zd.begin();zz!=zd.end();zz++)if (zz->second>zd[result])result=zz->first;return result;}
};

438. 找到字符串中所有字母异位词

class Solution {
public:vector<int> findAnagrams(string s, string p) {vector<int> lb;if (s.size()<p.size()) return lb;unordered_map<char,int> zd1,zd2;for (int i=0;i<p.size();i++){if (zd1.find(p[i])==zd1.end()) zd1[p[i]]=0;zd1[p[i]]+=1;if (zd2.find(s[i])==zd2.end()) zd2[s[i]]=0;zd2[s[i]]+=1;}if (zd1==zd2) lb.push_back(0);for (int i=1;i<s.size()-p.size()+1;i++){zd2[s[i-1]]-=1;if (zd2[s[i-1]]==0) zd2.erase(s[i-1]);if (zd2.find(s[i-1+p.size()])==zd2.end()) zd2[s[i-1+p.size()]]=0;zd2[s[i-1+p.size()]]+=1;if (zd1==zd2) lb.push_back(i);}return lb;}
};

就做到这里吧。