算法专题：滑动窗口
参考练习习题总集
文章目录
- 3. 无重复字符的最长子串
- 30. 串联所有单词的子串
- 76. 最小覆盖子串
- 187. 重复的DNA序列
- 219. 存在重复元素 II
- 220. 存在重复元素 III
- 396. 旋转函数
- 424. 替换后的最长重复字符
- 438. 找到字符串中所有字母异位词
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
这个确实牛，不看题解根本想不到。
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。