无重复字符的最长子串

class Solution {
    public int lengthOfLongestSubstring(String s) {
        Map<Character,Integer> map = new HashMap<>();
        int len = s.length();
        int ans = 0;
        char[] c = s.toCharArray();
        /*  c[]    p w w k e w
            key    0 0 0 0 0 0    
            value  0 0 0 0 0 0    ans=0 start=0 end=0

            key    p 0 0 0 0 0    p
            value  1 0 0 0 0 0    ans=1 start=0 end=0

            key    p w 0 0 0 0    p w
            value  1 2 0 0 0 0    ans=2 start=0 end=1

            key    p w 0 0 0 0    w
            value  1 3 0 0 0 0    ans=2 start=2 end=2

            key    p w k 0 0 0    w k
            value  1 3 4 0 0 0    ans=2 start=2 end=3

            key    p w k e 0 0    w k e
            value  1 3 4 5 0 0    ans=3 start=2 end=4

            key    p w k e 0 0    w k e w
            value  1 6 4 5 0 0    ans=3 start=3 end=5
        */
        for(int start = 0, end = 0; end < len; end++){
            if(map.containsKey(c[end])){
                start = Math.max(start, map.get(c[end]));
                //start = map.get(c[end]);  错误写法
            }
            map.put(c[end],end+1);
            ans = Math.max(ans, end-start+1);
        }
        return ans;
    }
    /*      c[]    a b b a
            key    0 0 0 0     
            value  0 0 0 0    ans=0 start=0 end=0

            key    a 0 0 0    a
            value  1 0 0 0    ans=1 start=0 end=0

            key    a b 0 0    a b
            value  1 2 0 0    ans=2 start=0 end=1

            key    a b 0 0    b 
            value  1 3 0 0    ans=2 start=2 end=2



            map.containsKey(c[end]) == true
            map.get(c[end]) -> map.get(c[3]) -> map.get('a') -> 1
            此时  map.get(c[end]) < start  注意这种情况，所以start = Math.max(start, map.get(c[end]));             
            要这样写，充分说明了不是所有情况下map.get(c[end])总是那个较大值。

            key    a b 0 0    ba 
            value  1 ? 0 0    ans=? start=? end=3
        */
}