字符串leetcode3，30，1234
作者：互联网
字符串
leetcode3.无重复字符的最长字串
public int lengthOfLongestSubstring(String s) {
//如果s为空，length不大于0，是一个空串，就没有向下执行的必要了
if(s != null && s.length() > 0 && s != ""){
//String -> char[]
char[] strChar = s.toCharArray();
// 存储最长字串 key:char值，value:index下标
ArrayList<String> maxStr = new ArrayList<>();
//临时的字串存储空间
ArrayList<String> tempStr = new ArrayList<>();
//循环
for(int i=0; i<strChar.length; i++){
//char -> String
String str = new String(new char[]{strChar[i]});
//判断str是否存在于tempStr中
if(tempStr.contains(str)){
//先判断tempStr的长度是否大于等于maxStr的长度,大于，才能将最长字串覆盖
if(tempStr.size() > maxStr.size()){
maxStr = new ArrayList<>(tempStr);
}
//存储重复字符
int reIndex = tempStr.indexOf(str);
// 删除tempStr中的重复字节及其之前的字符
for(int j=0;j<=reIndex;j++){
tempStr.remove(0);
}
}
//将当前字符存入tempStr中
tempStr.add(str);
}
//最终判断
if(tempStr.size() > maxStr.size()){
maxStr = tempStr;
}
//返回最长字串的长度
return maxStr.size();
}
return 0;
}
leetcode30.串联所有单词的子串
public static List<Integer> findSubstring(String s, String... words) {List<Integer> list = new ArrayList<>();
if (s.length() == 0 || words.length == 0)
return list;
int len = words[0].length();
if (s.length() < words.length * len)
return list;
int idx = -1;
char[] sCharS = s.toCharArray();
Set<Integer> allSet = new HashSet<Integer>();// 存储在字符串S所有匹配的下标 去重 排序
Map<String, Integer> wordMap = new HashMap<>();// 存储 word中各个单词的个数
Map<Integer, String> idxMap = new HashMap<>();// 存储字符串s各下标对应的单词
for (String word : words) {
if (wordMap.containsKey(word)) {
wordMap.put(word, wordMap.get(word) + 1);
continue;
}
char[] tem = word.toCharArray();
while ((idx = indexOf(sCharS, sCharS.length, tem, len, idx + 1)) > -1) {
idxMap.put(idx, word);
allSet.add(idx);
}
wordMap.put(word, 1);
}
String word;
int slideLen = len * words.length;// 滑块长度
int n;// 临时变量
Map<String, Integer> temWordMap = new HashMap<>();
for (int k = 0; k < len; k++) {
int flagNum = 0;// 表示滑块中有效的单词个数
for (int i = -1, j = k; j <= sCharS.length - len; j += len) {
if (allSet.contains(j)) {
if (i == -1) {// 初始化滑块
i = j;
flagNum = 0;
temWordMap.clear();
temWordMap.putAll(wordMap);
}
// 滑块长度增加 在尾部添加
word = idxMap.get(j);
n = temWordMap.get(word) - 1;
temWordMap.put(word, n);
if (n >= 0)
flagNum++;
if (j - i >= slideLen) {// 滑块长度减小 吐出头部数据
word = idxMap.get(i);
n = temWordMap.get(word) + 1;
temWordMap.put(word, n);
if (n > 0)
flagNum--;
i += len;
}
if (flagNum == words.length)
list.add(i);
} else {
i = -1;// j所在的位置不是给定的单词 ，销毁滑块
}
}
}
return list;
}
leetcode1234.替换子串得到平衡字符串
class Solution:
def balancedString(self, s: str) -> int:
n = len(s)
b = n // 4
from collections import Counter
counter = Counter(s)
counter = {key:value for key,value in counter.items() if value > b}
if not counter or n < 4:
return 0
rmove = True
l,r = 0,0
minlen = n
while l <= r and r < n:
if s[r] in counter and rmove:
counter[s[r]] -= 1
elif l > 0 and s[l - 1] in counter and not rmove:
counter[s[l - 1]] += 1
if {key:value for key,value in counter.items() if value > b}:
r += 1
rmove = True
else:
minlen = min(minlen, r - l + 1)
l += 1
rmove = False
return minlen
Guoliang Li
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标签：1234,word,leetcode3,int,30,length,tempStr,new,String 来源： https://blog.csdn.net/qq_35740095/article/details/103995060