LeetCode Longest Substring Without Repeating Characters
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<pre name="code" class="cpp">
class Solution {public: int hashtable[128]; int lengthOfLongestSubstring(string s) { int ans=0; memset(hashtable,0,sizeof(hashtable));//标记是否被访问过 for (int i=0; i<s.length(); i++) {//以s[i]为开头的子串 memset(hashtable, 0, sizeof(hashtable)); int len=0; //寻找以s[i]为开头的最长连续不重复子串 for (int j=i; j<s.length(); j++) { if (hashtable[s[j]]) { break; }else{ hashtable[s[j]]=1; len++; ans=max(len,ans); } } } return ans; } };
看到一个大神的解法。
class Solution {public: int lengthOfLongestSubstring(string s) { int ans = 0; int dic[256]; memset(dic,-1,sizeof(dic)); int len = s.size(); int idx = -1; for (int i=0;i<len;i++) { char c = s[i]; if (dic[c]>idx) idx = dic[c]; ans = max(ans,i-idx); dic[c] = i; } return ans; }};只要一遍遍历就可以求解。时间复杂度为O(n)。仔细看了看代码,我的理解是,每遍历到一个字符s[i]的时候,就去计算s[i]为结尾的最长不重复子串的长度。而以s[i]为结尾的最长子串的范围一定是[max(dic[s[i]],idx)+1,i],其中,dic[s[i]]是字符s[i]上一次出现的位置。idx是以s[i-1]为结尾的最长不重复子串的头字符的前一个字符下标。也就是说假如包含了idx,那么[idx,i-1]内就存在了重复字符,那么,[idx,i]也就存在了重复字符,假如包含了dic[s[i]]那么,dic[s[i]]所指向的字符与s[i]重复。因而,以s[i]为结尾的最长子串的范围应该是[max(dic[s[i]],idx),i]。其长度为i-(max([dic[s[i]]],idx)+1)+1,即为:i-max(dic[s[i]],idx)。然后与ans取max 更新ans,更新idx。
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