LeetCode003 Longest Substring Without Repeating Characters

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详细见：leetcode.com/problems/longest-substring-without-repeating-characters/

说明：map/dict/数组

Java Solution: github

package leetcode;import java.util.Arrays;public class P003_LongestSubstringWithoutRepeating {public static void main(String[] args) {new Solution1().lengthOfLongestSubstring("mabcdabcdmmm");new Solution1().lengthOfLongestSubstring("mabcd");}/* * 82.35% */static class Solution {int[] map = new int[128];    public int lengthOfLongestSubstring(String s) {        if (s == null)        return 0;        if (s.length() < 2)        return s.length();        int[] preIndex = new int[s.length()];        Arrays.fill(map, -1);        for (int i = 0; i != preIndex.length; i ++) {        int index = (int)s.charAt(i);        if (map[index] == -1) {        preIndex[i] = -1;        map[index] = i;        } else {        preIndex[i] = map[index];        map[index] = i;        }        }        int i0 = 0, i1 = 0, max = Integer.MIN_VALUE;        for (; i1 != preIndex.length; i1 ++) {        if (preIndex[i1] >= i0) {        if (max < i1 - i0)        max = i1 - i0;        i0 = preIndex[i1] + 1;        }        }    if (max < i1 - i0)    max = i1 - i0;        System.out.println(max);        return max;    }}/* * 将两次O(N)整合成一次O(N) * 86.95% */static class Solution1 {int[] map = new int[128];    public int lengthOfLongestSubstring(String s) {        if (s == null)        return 0;        int len = s.length();        if (len < 2)        return len;        Arrays.fill(map, -1);        int i0 = 0, i1 = 0, max = Integer.MIN_VALUE, preIndex = 0;        for (; i1 != len; i1 ++) {        int index = (int)s.charAt(i1);        if (map[index] == -1) {        preIndex = -1;        map[index] = i1;        } else {        preIndex = map[index];        map[index] = i1;        }        if (preIndex >= i0) {        if (max < i1 - i0)        max = i1 - i0;        i0 = preIndex + 1;        }        }    if (max < i1 - i0)    max = i1 - i0;        return max;    }}}

C Solution: github

/*    url: leetcode.com/problems/longest-substring-without-repeating-characters/    19ms 60.40%*/#include <stdio.h>#include <stdlib.h>#include <string.h>int lengthOfLongestSubstring(char* s) {    int m[128];    int i = 0, j = 0;    int n = 0, p = 0;    int * c = NULL;    int max = 0;    int v = 0;    if (s == NULL || * s == '\0')        return 0;    for (i = 0; i < 128; i ++)        m[i] = -1;    n = strlen(s);    c = (int *) malloc(sizeof(int) * n);    for (i = 0,j = 0; j < n; j ++) {        v = *(s + j);        p = m[v];        m[v] = j;        if (p >= i) {            max = max < j - i ? j - i : max;            i = p + 1;        }    }    max = max < j - i ? j - i : max;    free(c);    return max;}int main() {    char s[] = "pwwkew";    printf("answer is %d\r\n", lengthOfLongestSubstring(s));    return 0;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/longest-substring-without-repeating-characters/    dict保存上一次字符出现的位置，出现重复，从下一个开始计数    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年2月12日    @details:    Solution1: AC 102ms 71.02%    @details:    Solution1: AC 192ms 17.26%'''class Solution1(object):    def lengthOfLongestSubstring(self, s):        """        :type s: str        :rtype: int        """        if s == None or len(s) == 0: return 0        d,m,u = {},1,0        for i in range(len(s)):            p = d.get(s[i])            d[s[i]] = i            if p == None: p = -1            elif p >= u:                m = max(m, i - u)                u = p + 1        m = max(m, len(s) - u)        return mclass Solution2(object):    def lengthOfLongestSubstring(self, s):        """        :type s: str        :rtype: int        """        if s ==  None or len(s) == 0:            return 0        d, m , u = [-1] * 256, -1, -1        for j in range(len(s)):            p = d[ord(s[j])]            d[ord(s[j])] = j            m = max(m, j - u)            u = max(u, p + 1)        m = max(m, len(s) - u)            return m        if __name__ == "__main__":    d = "abba"    s = Solution2()    print(s.lengthOfLongestSubstring(d))

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