3. Longest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters.
Examples:
Given "abcabcbb", the answer is "abc", which the length is 3.
Given "bbbbb", the answer is "b", with the length of 1.
Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
解法1：遍历字符串找最长不重复字串，用hash表记录下已有的字符，用begin来记录开始计数的位置。如果下一个字符在hash表中已经记录了，就修改begin和hash的值来继续下一次查找
1 class Solution { 2 public: 3 int lengthOfLongestSubstring(string s) { 4 int maxLength=0;//记录下最长的不重复字串的长度 5 int length=0; 6 int begin=0;//记录下开始计算长度的位置 7 short hash[128];//哈希表，ASCII码，最多不超过128个字符 8 memset(hash, 0, sizeof(hash)); 9 for(int idx=0; idx<s.size(); ++idx) 10 { 11 12 int key=s[idx]-'\0'; 13 if(hash[key]==0) 14 { 15 hash[key]=1; 16 ++length; 17 }else{ 18 if(length>maxLength)maxLength=length; 19 if(maxLength==128)return 128;//如果最大长度达到128，立即返回，无须再比较下去 20 while(s[begin]!=s[idx]) 21 { 22 int key=s[begin]-'\0'; 23 hash[key]=0; 24 --length; 25 ++begin; 26 } 27 ++begin; 28 } 29 } 30 if(length>maxLength)maxLength=length;//别忘了最后一次更新 31 return maxLength; 32 } 33 };
解法2：动态规划，遍历字符串，在hash表中存储的是字符在字符串中的位置，时间复杂度为O(n)，比上面的算法更快
1 class Solution { 2 public: 3 int lengthOfLongestSubstring(string s) { 4 int maxLength=0; 5 int length=0; 6 int hash[128]; 7 for(int i=0; i<sizeof(hash)/4; ++i)hash[i]=-1;//初始化为-1 8 for(int idx=0; idx<s.size(); ++idx) 9 { 10 int val=hash[s[idx]-'\0']; 11 if(val==-1 || idx-length>val)//如果为-1，表示没有出现，故length加1。如果出现在计数范围之前，不影响，同没有出现 12 ++length; 13 else{//如果出现在计数范围之内，更新maxLength，重置length 14 if(length>maxLength)maxLength=length; 15 if(maxLength==128)return 128; 16 length=idx-val; 17 } 18 hash[s[idx]-'\0']=idx; 19 } 20 if(length>maxLength)maxLength=length;//最后一次更新maxLength 21 return maxLength; 22 } 23 };
