LeetCode 3.Longest Substring Without Repeating Characters

1.动态规划 54 ms
(1)last[] 存储当前字符的上一个相同字符的下标，-1表示在当前字符之前不存在相同的
如字符串abbac 对应的last为-1 -1 1 0 -1
(2)状态转移

    start[]存储以当前字符结尾的最长子串的开始位置，也就是说此时对应的字串应该为s(start[i]…i)

     对于当前位置i，
     如果last[i] < start[i-1],那么位置i-1对应的子串中并没有字符s[i]，此时是s[i]可以加入到i-1时的子串中，也就是说start[i] = start[i-1] + 1;
    如果last[i] >= start[i-1]， 那么子串s(start[i-1]…i-1)中是存在着s[i]字符的， 此时i对应的子串应该从last[i]+1开始，即重复字符的下一个字符

public class Solution {        public int lengthOfLongestSubstring(String s) {                     if (null == s || 0 == s.length()){                return 0;            }            char[] sequence = s.toCharArray();            int length = sequence.length;            //last[] 存储当前字符的上一个相同字符的位置            // abbac -1 -1 1 0 -1            int[] last = new int[length];            Map<Character,Integer> map = new HashMap<Character,Integer>();            for (int i = 0; i < length ; i++){                Character ch = sequence[i];                if (map.containsKey(ch)){                    last[i] = map.get(ch);                }else{                    last[i] = -1;                }                map.put(ch, i);            }            //start[]存储以当前字符结尾的最长子串的开始位置            //动态转移：            //if last[i] < start[i - 1],then start[i] = start[i - 1];            //if last[i] >= start[i - 1] and last[i] <= i-1,then start[i] = last[i] + 1;            int maxLen = 1;            int start[] = new int[length];                      start[0] = 0;            for (int i = 1; i < length; i++){                int start_i_1 = start[i-1];                int last_i = last[i];                if (last_i < start_i_1){                    start[i] = start_i_1;                }else{                    start[i] = last_i + 1;                }                maxLen = Math.max(maxLen, 1+ i - start[i]);            }            return maxLen;        }    }

2.优化 36ms

    public class Solution {        public int lengthOfLongestSubstring(String s) {            if (null == s || 0 == s.length()) {                return 0;            }            char[] sequence = s.toCharArray();            int length = sequence.length;            int last, pre = 0;            int maxLen = 1;            int[] map = new int[128];            for (int i = 0; i < length; i++) {                Character ch = sequence[i];                last = map[(int) ch] - 1;                map[(int) ch] = i + 1;                pre = (last < pre) ? pre : last + 1;                maxLen = maxLen > i - pre + 1 ? maxLen :  i - pre + 1 ;            }            return maxLen;        }    }