五个问题，一次解决，字符子串问题总结
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我发现在 leetcode 的问题中，至多有 5 个子字符串寻找问题能够用滑动窗口算法解决，因而我在这里总结了这类算法的模版，心愿能够帮忙你。
1）模版
public class Solution {public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
// 依据问题，初始化一个贮存后果的容器
List<Integer> result = new LinkedList<>();
if (t.length()> s.length()) return result;
// 创立一个 hashmap 来保留指标子串中的字符
//(K, V) = (Character, Frequence of the Characters)
//key 是字符，value 是该字符呈现的次数
Map<Character, Integer> map = new HashMap<>();
// 将指标子串转为 map 存储
for(char c : t.toCharArray()){map.put(c, map.getOrDefault(c, 0) + 1);
}
// 保护一个计数器，去查看是否匹配指标字符串
int counter = map.size();// 必须是 map 的长度，不是字符串的长度是因为可能元素有反复。// 两个点，窗口的左端点和右端点
int begin = 0, end = 0;
// 匹配指标字符串的子字符串的长度
int len = Integer.MAX_VALUE;
// 从源字符串循环
while (end < s.length()) {char c = s.charAt(end);// 失去右端点处的字符
if (map.containsKey(c)) {map.put(c, map.get(c)-1);// 加一或减一
if (map.get(c) == 0) counter--;// 依据不同的条件批改计数器
}
end++;
//increase begin pointer to make it invalid/valid again
// 计数器条件：不同的问题抉择不同的条件
while (counter == 0) {char tempc = s.charAt(begin);// 留神：抉择字符是在开始端点而不是完结端点
if (map.containsKey(tempc)) {map.put(tempc, map.get(tempc) + 1);// 加减一
if (map.get(tempc) > 0) counter++;// 依据不同的需要批改计数器
}
/* save / update(min/max) the result if find a target*/
// 如果发现一个指标，保留、更新（最小、最大）后果
// result collections or result int value
begin++;
}
}
return result;
}
}2）相干问题
https://leetcode-cn.com/probl…
https://leetcode-cn.com/probl…
https://leetcode-cn.com/probl…
https://leetcode-cn.com/probl…
https://leetcode-cn.com/probl…
3）具体问题如何利用模版
438. 找到字符串中所有字母异位词
https://leetcode-cn.com/probl…
public class Solution {public List<Integer> findAnagrams(String s, String t) {List<Integer> result = new LinkedList<>();
if(t.length()> s.length()) return result;
Map<Character, Integer> map = new HashMap<>();
for(char c : t.toCharArray()){map.put(c, map.getOrDefault(c, 0) + 1);
}
int counter = map.size();
int begin = 0, end = 0;
int head = 0;
int len = Integer.MAX_VALUE;
while (end < s.length()) {char c = s.charAt(end);
if (map.containsKey(c)) {map.put(c, map.get(c) - 1);
if (map.get(c) == 0) counter--;
}
end++;
//counter 等于 0 意味着，end 之前至多有可能凑出 target 的字母数量
while (counter == 0) {char tempc = s.charAt(begin);
if (map.containsKey(tempc)) {map.put(tempc, map.get(tempc) + 1);
if(map.get(tempc) > 0){counter++;}
}
if (end - begin == t.length()) {result.add(begin);
}
begin++;
}
}
return result;
}
}76. 最小笼罩子串
https://leetcode-cn.com/probl…
public class Solution {public String minWindow(String s, String t) {if(t.length()> s.length()) return "";
Map<Character, Integer> map = new HashMap<>();
for(char c : t.toCharArray()){map.put(c, map.getOrDefault(c,0) + 1);
}
int counter = map.size();
int begin = 0, end = 0;
int head = 0;
int len = Integer.MAX_VALUE;
while(end < s.length()){char c = s.charAt(end);
if(map.containsKey(c) ){map.put(c, map.get(c)-1);
if(map.get(c) == 0) counter--;
}
end++;
while(counter == 0){char tempc = s.charAt(begin);
if(map.containsKey(tempc)){map.put(tempc, map.get(tempc) + 1);
if(map.get(tempc) > 0){counter++;}
}
if(end-begin < len){
len = end - begin;
head = begin;
}
begin++;
}
}
if(len == Integer.MAX_VALUE) return "";
return s.substring(head, head+len);
}
}3. 无反复字符的最长子串
https://leetcode-cn.com/probl…
public class Solution {public int lengthOfLongestSubstring(String s) {Map<Character, Integer> map = new HashMap<>();
int begin = 0, end = 0, counter = 0, d = 0;
while (end < s.length()) {
// > 0 means repeating character
//if(map[s.charAt(end++)]-- > 0) counter++;
char c = s.charAt(end);
map.put(c, map.getOrDefault(c, 0) + 1);
if(map.get(c) > 1) counter++;
end++;
while (counter > 0) {//if (map[s.charAt(begin++)]-- > 1) counter--;
char charTemp = s.charAt(begin);
if (map.get(charTemp) > 1) counter--;
map.put(charTemp, map.get(charTemp)-1);
begin++;
}
d = Math.max(d, end - begin);
}
return d;
}
}30. 串联所有单词的子串
https://leetcode-cn.com/probl…
public class Solution {public List<Integer> findSubstring(String S, String[] L) {List<Integer> res = new LinkedList<>();
if (L.length == 0 || S.length() < L.length * L[0].length()) return res;
int N = S.length();
int M = L.length; // *** length
int wl = L[0].length();
Map<String, Integer> map = new HashMap<>(), curMap = new HashMap<>();
for (String s : L) {if (map.containsKey(s)) map.put(s, map.get(s) + 1);
else map.put(s, 1);
}
String str = null, tmp = null;
for (int i = 0; i < wl; i++) {
int count = 0; // remark: reset count
int start = i;
for (int r = i; r + wl <= N; r += wl) {str = S.substring(r, r + wl);
if (map.containsKey(str)) {if (curMap.containsKey(str)) curMap.put(str, curMap.get(str) + 1);
else curMap.put(str, 1);
if (curMap.get(str) <= map.get(str)) count++;
while (curMap.get(str) > map.get(str)) {tmp = S.substring(start, start + wl);
curMap.put(tmp, curMap.get(tmp) - 1);
start += wl;
//the same as https://leetcode.com/problems/longest-substring-without-repeating-characters/
if (curMap.get(tmp) < map.get(tmp)) count--;
}
if (count == M) {res.add(start);
tmp = S.substring(start, start + wl);
curMap.put(tmp, curMap.get(tmp) - 1);
start += wl;
count--;
}
}else {curMap.clear();
count = 0;
start = r + wl;//not contain, so move the start
}
}
curMap.clear();}
return res;
}
}