算法第四周作业01

Description

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

Solution

1.  采用动态规划法；

2. 指定dp[i]数组为第i个字符结尾的最大不重复子字符串，即对于"abcabcbb"，dp = {1,2,3,3,3,3,2,1};

3. dp[i]可以通过dp[i-1]来获得，例如当第(i-dp[i-1])到第(i-1)字符均与第i个字符不一致，则dp[i] = dp[i-1] + 1;

否则找到与第i个字符一直的元素的索引index，dp[i] = i-index;

Code

public int lengthOfLongestSubstring(String s) { int len = s.length(); // 如果长度为0则直接返回 if (len == 0) { return 0; } // 设置dp int[] dp = new int[len]; // 转成char，方便比较 char[] arr = s.toCharArray(); // 初始化dp[0] dp[0] = 1; int max = 1; // 对dp的值逐个获取 for (int i = 1; i < len; i++) { int j = i - 1; // 与之前的子字符串进行比较 for (; j >= i - dp[i - 1]; j--) { if (arr[j] == arr[i]) { break; } } dp[i] = i - j; // 如果比之前所有最大子字符串还要大，则替换 if (max < dp[i]) { max = dp[i]; } } return max; }