题目意思:求字符串中,最长不重复连续子串

思路:使用hashmap,发现unordered_map会比map快,设置一个起始位置,计算长度时,去减起始位置的值

   eg:a,b,c,d,e,c,b,a,e

0 1 2 3 4

0 1 5 3 4

0 6 5 3 4

7 6 5 3 4

7 6 5 3 8

     再次出现c时,将start值置为map[c]+1=3,对于下一个b,因为map[b]<start,则直接map[b]=i

 class Solution {
public:
int lengthOfLongestSubstring(string s) {
unordered_map<char,int> mymap;
int flag=;
int start=,i=;
unordered_map<char,int>::iterator ite;
while(i<s.size()){
ite=mymap.find(s[i]);
if(ite==mymap.end()){
mymap[s[i]]=i;
flag=max(flag,i-start+);
}
else{
if(mymap[s[i]]>=start)
start=mymap[s[i]]+;
mymap[s[i]]=i;
flag=max(flag,i-start+);
}
++i;
}
return flag;
}
};

时间复杂度:O(n)

超时解法:

 class Solution {
public:
int lengthOfLongestSubstring(string s) {
map<char,int> mymap;
int i=,max=,flag;
map<char,int>::iterator ite;
while(i<s.length()){
ite=mymap.find(s[i]);
if(ite==mymap.end()){
mymap[s[i]]=i;
if(mymap.size()>max)
max=mymap.size();
}
else{
flag=ite->second;
mymap.clear(); //每次清空,是及其失败的做法
for(int j=flag+;j<=i;++j){
mymap[s[j]]=j;
}
}
++i;
}
return max;
}
};

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