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.
思路:动态规划.
dp[i]表示以 i 结尾的最长无重复子串的长度.从前向后不断更新dp[j].
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int len = s.length(), mmax = 1;
if(len == 0)
{
return 0;
}
int dp[100000];
dp[0] = 1;
for(int i=1;i<len;i++)
{
int maxlen = dp[i-1];//最多向前面找 dp[i-1]个
int tot = 1;
for(int j=0;j<maxlen;j++)
{
if(s[i-j-1] != s[i])
tot++;
else
break;
}
if(tot > mmax)
mmax = tot;
dp[i] = tot;
}
return mmax;
}
};

本文介绍了一种使用动态规划求解给定字符串中最长无重复字符子串长度的方法。通过示例说明了如何实现这一算法,包括针对不同输入情况的处理方式。
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