[LeetCode] 003. Longest Substring Without Repeating Characters (Medium) (C++/Java/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

003.Longest_Substring_Without_Repeating_Characters (Medium)

链接：

题目：https://oj.leetcode.com/problems/Longest-Substring-Without-Repeating-Characters/
代码(github)：https://github.com/illuz/leetcode

题意：

从标题就可以知道题意了，是求一个字符串中最长的不含重复字符的子串。

分析：

开一个数组记录当前字符最近出现的位置，一遍算过去，更新左边界，用它计算最大值就行了。
需要花费常数的空间。

代码：

C++:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
       int maxlen = 0, left = 0;
	   int sz = s.length();
	   int prev[N];
	   memset(prev, -1, sizeof(prev));

	   for (int i = 0; i < sz; i++) {
		   if (prev[s[i]] >= left) {
			   left = prev[s[i]] + 1;
		   }
		   prev[s[i]] = i;
		   maxlen = max(maxlen, i - left + 1);
	   }
	   return maxlen;
    }
};

Java:

public class Solution {

    public int lengthOfLongestSubstring(String s) {
        int res = 0, left = 0;
        int prev[] = new int[300];

		// init prev array
        for (int i = 0; i < 300; ++i)
            prev[i] = -1;

        for (int i = 0; i < s.length(); ++i) {
            if (prev[s.charAt(i)] >= left)
                left = prev[s.charAt(i)] + 1;
            prev[s.charAt(i)] = i;
            if (res < i - left + 1)
                res = i - left + 1;
        }
        return res;
    }
}

Python:

class Solution:
    # @return an integer
    def lengthOfLongestSubstring(self, s):
        res = 0
        left = 0
        d = {}

        for i, ch in enumerate(s):
            if ch in d and d[ch] >= left:
                left = d[ch] + 1
            d[ch] = i
            res = max(res, i - left + 1)
        return res