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.
public class Solution {
public int lengthOfLongestSubstring(String s) {
Map<Character,Integer> map=new HashMap<Character,Integer>();
int max=0;
for(int i=0,j=0;i<s.length();i++){
if(map.containsKey(s.charAt(i))){
j=Math.max(j,map.get(s.charAt(i))+1);
}
map.put(s.charAt(i),i);
max=Math.max(max,i-j+1);
}
return max;
}
}思路:
map中存放字符以及字符的位置,指针j用途是记录当前字符串的起始位置,更新j方法是,重复字符的位置+1与当前j比较,这样排可以排除当前字符与j之前的字符重复的情况。遍历字符,判断map是否包含字符,如果包含则更新j,然后更新map 和max。
本文介绍了一种求解字符串中最长无重复字符子串长度的高效算法,并提供了详细的实现思路及Java代码示例。
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