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五個(gè)問題丛版，一次解決穿剖，字符子串問題總結(jié)
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我發(fā)現(xiàn)在leetcode的問題中哲思，至少有5個(gè)子字符串尋找問題可以用滑動窗口算法解決惯殊，因此我在這里總結(jié)了這類算法的模版，希望可以幫助你也殖。
1）模版
public class Solution {
public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
//根據(jù)問題土思，初始化一個(gè)儲存結(jié)果的容器
List<Integer> result = new LinkedList<>();
if (t.length()> s.length()) return result;
//創(chuàng)建一個(gè)hashmap來保存目標(biāo)子串中的字符
//(K, V) = (Character, Frequence of the Characters)
//key是字符， value是該字符出現(xiàn)的次數(shù)
Map<Character, Integer> map = new HashMap<>();
//將目標(biāo)子串轉(zhuǎn)為map存儲
for(char c : t.toCharArray()){
map.put(c, map.getOrDefault(c, 0) + 1);
}
//維護(hù)一個(gè)計(jì)數(shù)器忆嗜，去檢查是否匹配目標(biāo)字符串
int counter = map.size();//必須是map的長度己儒，不是字符串的長度是因?yàn)榭赡茉赜兄貜?fù)。
//兩個(gè)點(diǎn)捆毫，窗口的左端點(diǎn)和右端點(diǎn)
int begin = 0, end = 0;
//匹配目標(biāo)字符串的子字符串的長度
int len = Integer.MAX_VALUE;
//從源字符串循環(huán)
while (end < s.length()) {
char c = s.charAt(end);//得到右端點(diǎn)處的字符
if (map.containsKey(c)) {
map.put(c, map.get(c)-1);//加一或減一
if (map.get(c) == 0) counter--;//根據(jù)不同的條件修改計(jì)數(shù)器
}
end++;
//increase begin pointer to make it invalid/valid again
//計(jì)數(shù)器條件：不同的問題選擇不同的條件
while (counter == 0) {
char tempc = s.charAt(begin);//注意：選擇字符是在開始端點(diǎn)而不是結(jié)束端點(diǎn)
if (map.containsKey(tempc)) {
map.put(tempc, map.get(tempc) + 1);//加減一
if (map.get(tempc) > 0) counter++;//根據(jù)不同的需求修改計(jì)數(shù)器
}
/* save / update(min/max) the result if find a target*/
//如果發(fā)現(xiàn)一個(gè)目標(biāo)闪湾，保存、更新（最小绩卤、最大）結(jié)果
// result collections or result int value
begin++;
}
}
return result;
}
}
2）相關(guān)問題
https://leetcode-cn.com/problems/minimum-window-substring/
https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
https://leetcode-cn.com/problems/substring-with-concatenation-of-all-words/
https://leetcode-cn.com/problems/longest-substring-with-at-most-two-distinct-characters/
https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/
3）具體問題如何應(yīng)用模版
438.找到字符串中所有字母異位詞
https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/
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意味著途样，end之前至少有能夠湊出target的字母數(shù)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. 最小覆蓋子串
https://leetcode-cn.com/problems/minimum-window-substring/
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.無重復(fù)字符的最長子串
https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
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.串聯(lián)所有單詞的子串
https://leetcode-cn.com/problems/substring-with-concatenation-of-all-words/
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;
}
}
