資訊專欄INFORMATION COLUMN
摘要：建立數(shù)組，存儲(chǔ)個(gè)字符最近一次出現(xiàn)的位置。首次出現(xiàn)某字符時(shí)，其位置標(biāo)記為，并用無重復(fù)字符計(jì)數(shù)器記錄無重復(fù)字符的長度，再在更新其最大值。循環(huán)完整個(gè)字符串后，返回最大值。
Problem
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.
Note建立int數(shù)組ch[]，存儲(chǔ)256個(gè)字符最近一次出現(xiàn)的位置。首次出現(xiàn)某字符a時(shí)，其位置標(biāo)記為ch[a]，并用無重復(fù)字符計(jì)數(shù)器count記錄無重復(fù)字符的長度，再在max更新其最大值。
當(dāng)出現(xiàn)重復(fù)字符a時(shí)，從無重復(fù)字符計(jì)數(shù)器count中減去兩個(gè)重復(fù)字符之間的長度(ch[a]-start)，并更新start為最后一個(gè)（任意）重復(fù)字符上一次出現(xiàn)的位置，然后更新重復(fù)字符的位置ch[a]。
循環(huán)完整個(gè)字符串后，返回最大值max。
舉例說明：
"pwwkewk" start = 0, count = 0, max = 0
i = 0:
a = "p" ch["p"] = 0 + 1 = 1 count = 1 max = 1
i = 1:
a = "w" ch["w"] = i + 1 = 2 count = 2 max = 2
i = 2:
a = "w"
ch["w"] = 2 > start = 0
count = count - (ch["w"] - start) = 2 - (2 - 0) = 0
start = 2
ch["w"] = 2 + 1 = 3
count = 1
max = 2
i = 3:
a = "k" ch["k"] = 3 + 1 = 4 count = 2 max = 2
i = 4:
a = "e" ch["e"] = 5 count = 3 max = 3
i = 5:
a = "w"
ch["w"] = 3 > start = 2
count = count - (ch["w"] - start) = 3 - (3 - 2) = 2
start = 3
ch["w"] = 6
count = 2 + 1 = 3
max = 3
i = 6:
a = "k"
ch["k"] = 4
start = 3
count = count - (ch["k"] - start) = 3 - (4 - 3) = 2
start = 4
ch["k"] = 7
count = 3
max = 3
Solution
public class Solution {
public int lengthOfLongestSubstring(String s) {
int[] ch = new int[256];
int max = 0, count = 0, start = 0;
for (int i = 0; i < s.length(); i++) {
char a = s.charAt(i);
if (ch[a] > start) {
count -= ch[a] - start;
start = ch[a];
}
ch[a] = i+1;
max = Math.max(max, ++count);
}
return max;
}
}
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