3. Longest Substring Without Repeating Characters
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/*
O(N)的算法,遍历一遍字符串,记录当前处理到的位置和每个字符最后出现的位置,然后更新当前的最长子串即可
比较好的一道题,思路需要好好记住
*/
class Solution {
public:int lengthOfLongestSubstring(string s) {
int flag[200]; //记录到当前位置之前字符s[i]出现的最后位置
int cur=0; //表示当前处理到的位置
int max=0; //记录每个位置当前的最大子串长度
for(int i=0;i<200;i++) flag[i]=0;
int tmp=0;
//每个字符的位置从1计数
for(int i=0;i<s.size();i++){
//如果上次s[i]字符出现的位置在当前处理位置(cur)之后就更新cur
if(flag[s[i]]>cur){
cur = flag[s[i]];
}
tmp = i+1-cur;
flag[s[i]] = i+1;
max = max<tmp?tmp:max;
}
return max;
}
};
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