golang_longest_substring_without_repeating_characters
Given a string s, find the length of the longest substring
without repeating characters.
Examples
Example 1:
Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
Constraints:
0 <= s.length <= 5 * 104sconsists of English letters, digits, symbols and spaces.
解析
給定一個字串 s
要求寫一個演算法找出子字串中不包含重複字元的最長長度
最直接的想法是
當還沒遇到重複字元時就累加長度。遇到重複字元時，就從上一次開始累加的位置start 右移一位， 也就是start+1 當作下一個開始做累計
具體作法如下：
初始化最大長度 maxLen = 0, start = 0, visitMap
然後逐步檢查每個字元 s[i]
每次先檢查 s[i] 是否已經存在 visitMap
如果s[i] 存在 visitMap 且 visitMap[s[i]] ≥ start 更新 start = vistMap[s[i]] +1
如果s[i] 不存在 visitMap ，更新 visitMap[s[i]] = i
更新 maxLen = max(maxLen, i - start +1)
當 start + maxLen ≥ len(s) 代表已經沒有辦法再找到更長不重複字元的子字串，直接回傳 maxLen
最後回傳 maxLen
程式碼
package sol
func lengthOfLongestSubstring(s string) int {
sLen := len(s)
start, maxLen := 0, 0
visitMap := make(map[byte]int)
var max = func(a, b int) int {
if a > b {
return a
}
return b
}
for end := 0; end < sLen; end++ {
// after start visit same character
if pos, ok := visitMap[s[end]]; ok && pos >= start {
start = pos + 1
}
visitMap[s[end]] = end
maxLen = max(maxLen, end-start+1)
if start+maxLen >= sLen {
break
}
}
return maxLen
}困難點
- 要看出找出不重複字元子字串每個位置間的關係
- 要理解需要透過 hashTable 去儲存當下遇到每個字元的最新位置，用來做遇到重複字元開始位置更新
Solve Point
- 初始化 start = 0, maxLen = 0, visitMap
- 從 i = 0.. len(s)-1 每次做以下運算
- 檢查 s[i] 是否有在 visitMap 中
- 如果有 , 檢查 visitMap[s[i]] ≥ start ， 更新 start = visitMap[s[i]]+ 1
- 更新 visitMap[s[i]] = i
- 更新 maxLen = max(maxLen, i - start +1)
- 當 start + maxLen ≥ len(s) 代表無法從 start 開始找到比 maxLen 長的不重複字元子字串 直接回傳 maxLen
- 回傳 maxLen