LeetCode 3. Longest Substring Without Repeating Characters (C++/python)

第二次

C++：

class Solution {public:    int lengthOfLongestSubstring(string s) {int length = 0;string subs;for (char c: s) {if (subs.find(c) == string::npos) {subs += c;length = max(length, int(subs.size()));} elsesubs = subs.substr(subs.find(c)+1) + c;}return length;}};

python:

class Solution(object):def lengthOfLongestSubstring(self, s):substr = ''length = 0i = 0for i in range(0, len(s)):if substr.find(s[i]) == -1:substr = substr + s[i]length = max(length, len(substr))else:substr = substr[substr.find(s[i])+1:] + s[i]return length

第一次

dp[i]代表自下标j (j<=i)起，到i的Longest Substring Without Repeating Characters.

已知dp[i-1], 则从i-1开始，往前遍历至i-dp[i-1], 看是否有重复的元素(下标为j)；

若有，dp[i] = i-j;

反之，dp[i] = dp[i-1] + 1;

for (int j = i-1; j>=i-dp[i-1]; -- j){if (s[i] == s[j]){dp[i] = i-j;break;}}// dp[i]初始化为0, 此时仍为0则表示未发现其与之前的[i-dp[i-1], i-1]这条子串中有相同的元素，即if没被执行dp[i] = dp[i]==0? dp[i-1]+1: dp[i]; 

代码：

class Solution {public:    int lengthOfLongestSubstring(string s)     {if (s.size()==0){return 0;}vector<int> dp(s.size(), 0);int max=1;dp[0] = 1;for (int i = 1; i < s.size(); ++ i){for (int j = i-1; j>=i-dp[i-1]; -- j){if (s[i] == s[j]){dp[i] = i-j;break;}}dp[i] = dp[i]==0? dp[i-1]+1: dp[i];max = dp[i]>max? dp[i]: max;}return max;    }};