leetcode 3. Longest Substring Without Repeating Characters
2017-05-07 20:52
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Given a string, find the length of the longest substring without repeating characters.
Examples:
Given
which the length is 3.
Given
with the length of 1.
Given
with the length of 3. Note that the answer must be a substring,
a subsequence and not a substring.
解:需要采用在线处理的方法,同时还要记录没哟个字符是否读取过,这可以使用map,其实看他人的solution也可以vector容器,都是同一种思路。
代码:class Solution {
public:
int lengthOfLongestSubstring(string s) {
map<int, int> hashMap;
int len = s.length();
int head = -1;
int maxlen = 0;
for(int i = 0; i < len; i++){
hashMap[s[i]] = -1;
}
for(int i = 0; i < len; i++){
if(hashMap[s[i]] > head){
head = hashMap[s[i]];
}
hashMap[s[i]] = i;
maxlen = max(maxlen, i - head);
}
return maxlen;
}
};
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:需要采用在线处理的方法,同时还要记录没哟个字符是否读取过,这可以使用map,其实看他人的solution也可以vector容器,都是同一种思路。
代码:class Solution {
public:
int lengthOfLongestSubstring(string s) {
map<int, int> hashMap;
int len = s.length();
int head = -1;
int maxlen = 0;
for(int i = 0; i < len; i++){
hashMap[s[i]] = -1;
}
for(int i = 0; i < len; i++){
if(hashMap[s[i]] > head){
head = hashMap[s[i]];
}
hashMap[s[i]] = i;
maxlen = max(maxlen, i - head);
}
return maxlen;
}
};
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