[leetcode]LeetCode刷题的日子--Longest Substring Without Repeating

题目：

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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.

题目翻译：

查找一个字符串中，最长不含有重复字母的字符串长度

题目思路：

创建一个字典，和一个起始标志。字典用于存储字母出现的位置，在我们遍历字符串的时候，如果在字母在字典中存储的位置小于起始标志，则认为从起始处开始，未出现重复字幕；否则认为出现重复情况，这个时候移动起始标志到当前位置，然后从起始位置处，继续开始遍历。我们遍历的位置，与起始标志的位置距离，就是不包含重复字符的字符串长度，在遍历过程中，找到最大值即可，注意边界条件。

代码：

use std::cmp;

fn length_of_longest_substring(s: String) -> i32 {
    let mut ret: i32 = 0;
    let mut start: i32 = -1;

    let mut dic = vec![-1; 128];

    let b = s.as_bytes();
    for i in (0..s.len()) {
        // if repeat just move the start
        if (dic[b[i as usize] as usize] > start) {
            start = dic[b[i as usize] as usize]
        }

        ret = std::cmp::max(ret, i as i32 - start);
        dic[b[i as usize] as usize] = i as i32;
    }

    return ret as i32;
}

#[test]
fn test_length_of_longest_substring() {
    let case1 = String::from("abcabcbb");
    let case2 = String::from("bbbbb");
    let case3 = String::from("pwwkew");

    assert_eq!(length_of_longest_substring(case1), 3);
    assert_eq!(length_of_longest_substring(case2), 1);
    assert_eq!(length_of_longest_substring(case3), 3);
}