Longest Substring Without Repeating Characters 字符串中最长的无重复子串长度
2014-12-04 20:49
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Longest Substring Without Repeating Characters
Given
a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
题目的意思是求字符串中最长的无重复子串长度,如上面的
abcabcbb 最长无重复子串为abc 长度为3
直接暴力法:代码如下
暴力法虽然简单,但时间过不了,下面用一个复杂度为n的算法
思路:
用一个256位数组asc[](字符共有256个,包括128个扩展字符),字符串中每一个字符的ASCII值对应该字符在数组中的位置
如字符a
在 asc[a]中的下标为97
初始化数组asc,令每一个元素为-1。定义start,start表示每次无重复字符串的其实点,初始值为0.从i=0开始遍历字符串,如果数组asc中没有出现该字符(asc[ch]==-1),那么令asc[ch]==i,记录下来该字符在字符串中的位置,接着遍历,如果遇到asc[ch]!=-1表示该字符已经出现了,需要将start到j的asc数组中的元素抹掉。然后重置start,令start=max(asc[i]+1,start);然后将
最后结果为 result=max(result,i-start);
代码如下:
Given
a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
题目的意思是求字符串中最长的无重复子串长度,如上面的
abcabcbb 最长无重复子串为abc 长度为3
直接暴力法:代码如下
class Solution { public: int lengthOfLongestSubstring(string s) { int i,j,temp,result=1; for(i=0;i<s.length();i++) { temp=1; for(j=0;j<s.length()-i;j++) { if(check(s,i,j)) temp++; else continue; } result=max(result,temp); } return result; } bool check(string s,int i,int j) { char ch=s[j]; for(int t=i;t<j;t++) { if(ch==s[t]) { return false; } } return true; } };
暴力法虽然简单,但时间过不了,下面用一个复杂度为n的算法
思路:
用一个256位数组asc[](字符共有256个,包括128个扩展字符),字符串中每一个字符的ASCII值对应该字符在数组中的位置
如字符a
在 asc[a]中的下标为97
初始化数组asc,令每一个元素为-1。定义start,start表示每次无重复字符串的其实点,初始值为0.从i=0开始遍历字符串,如果数组asc中没有出现该字符(asc[ch]==-1),那么令asc[ch]==i,记录下来该字符在字符串中的位置,接着遍历,如果遇到asc[ch]!=-1表示该字符已经出现了,需要将start到j的asc数组中的元素抹掉。然后重置start,令start=max(asc[i]+1,start);然后将
最后结果为 result=max(result,i-start);
代码如下:
<span style="font-size:18px;">class Solution { public: int lengthOfLongestSubstring(string s) { int result=0; int a[128]; int i,j,t,start=0; for(i=0;i<128;i++) a[i]=-1; for(i=0;i<s.length();i++) { if(a[s[i]]!=-1) { if(result<i-start) result=i-start; for(j=start;j<a[s[i]];j++) a[j]=-1; start=max(start,a[s[i]]+1); // result=max(result,i-start); } a[s[i]]=i; } result=max(result,i-start); return result; } };</span>
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