题目:
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
解题分析:
动态规划解题,分别计算前n个字符中以最后一个字符结尾的最长不重复子串,最后计算一下其中的最大值,时间复杂度为O(n*n+n)=O(n^2)。
代码:
public int lengthOfLongestSubstring(String s) {
if(s.length() == 0){
return 0;
}
int[] A = new int[s.length()];
A[0] = 1;
int max = 1;
for(int i = 1; i< s.length(); i++){
if(s.charAt(i) == s.charAt(i-1)){
A[i] = 1;
}else{
int j = 1;
for(; j < A[i-1]+1; j++){
if(s.charAt(i) == s.charAt(i-j)){
break;
}
}
A[i] = j;
}
if(max < A[i]){
max = A[i];
}
}
return max;
}
本文介绍了一种通过动态规划解决字符串中最长无重复子串问题的方法,并提供了完整的代码实现。该算法的时间复杂度为O(n^2),适用于寻找给定字符串中的最长非重复子串。
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