[LeetCode] Longest Substring Without Repeating Cha

CoderStudy 發(fā)布于2019-08-14 15:48 / 2355人閱讀

摘要：建立數(shù)組，存儲個字符最近一次出現(xiàn)的位置。首次出現(xiàn)某字符時，其位置標記為，并用無重復字符計數(shù)器記錄無重復字符的長度，再在更新其最大值。循環(huán)完整個字符串后，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256個字符最近一次出現(xiàn)的位置。首次出現(xiàn)某字符a時，其位置標記為ch[a]，并用無重復字符計數(shù)器count記錄無重復字符的長度，再在max更新其最大值。

當出現(xiàn)重復字符a時，從無重復字符計數(shù)器count中減去兩個重復字符之間的長度(ch[a]-start)，并更新start為最后一個（任意）重復字符上一次出現(xiàn)的位置，然后更新重復字符的位置ch[a]。

循環(huán)完整個字符串后，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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