LeetCode Top 100 Liked Questions 438. Find All Anagrams in a String (Java版; Medium)
题目描述
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"
Output:
[0, 6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:
s: "abab" p: "ab"
Output:
[0, 1, 2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-all-anagrams-in-a-string
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> list = new ArrayList<>();
HashMap<Character, Integer> map = new HashMap<>();
for(int i=0; i<p.length(); i++){
char ch = p.charAt(i);
map.put(ch, map.getOrDefault(ch, 0)-1);
}
int left=0, right=0;
int count = 0;
while(right<s.length()){
char ch = s.charAt(right);
if(map.containsKey(ch)){
map.put(ch, map.get(ch)+1);
if(map.get(ch)==0){
count++;
}
}
while(count==map.size()){
if(right - left + 1 == p.length()){
list.add(left);
}
char ch2 = s.charAt(left);
if(map.containsKey(ch2)){
map.put(ch2, map.get(ch2)-1);
if(map.get(ch2)==-1){
count--;
}
}
left++;
}
right++;
}
return list;
}
}第一次做; 双指针法:先移动right,满足条件后,再移动left; count变量记录有多少个字符满足数量要求了; 判断两个Integer类型的变量是否相等,得用equals()方法!! 不能用==, 但是可以用<或者>
import java.util.ArrayList;
import java.util.HashMap;
class Solution {
public List<Integer> findAnagrams(String s, String p) {
ArrayList<Integer> res = new ArrayList<>();
if(s.length() < p.length())
return res;
//
HashMap<Character, Integer> needs = new HashMap<>();
HashMap<Character, Integer> window = new HashMap<>();
//核心:记录window中有多少个符合要求的字符, 这个变量实现了O(1)是否window是否有效,最开始没有想到
int count = 0;
//创建needs
for(int i=0; i<p.length(); i++)
needs.put(p.charAt(i), needs.getOrDefault(p.charAt(i),0)+1);
//双指针主体
int left=0, right=0;
while(right<s.length()){
char ch = s.charAt(right);
if(needs.containsKey(ch)){
window.put(ch, window.getOrDefault(ch,0)+1);
//核心:检查ch的数量是否符合要求;
//细节: 满足条件后,count只会++一次
//顶级核心:哈希表中的value是Integer类型的,所以这里不能用window.get(ch)==needs.get(ch), 必须用.equals()方法
if(window.get(ch).equals(needs.get(ch))){
count++;
}
}
right++;
//核心:while循环条件
while(count==needs.size()){
//核心:if条件
if(right - left == p.length())
res.add(left);
char ch2 = s.charAt(left);
if(window.containsKey(ch2)){
window.put(ch2, window.get(ch2)-1);
//核心:检查ch2的数量是否不再符合要求;
/*
错误的想法: 不能用if(window.get(ch2) < needs.get(ch2)) 这样可能导致count执行多次减减,
不对, 实际上可以这么写, 因为count减减后就不满足下一轮循环的执行条件了!
*/
//细节: 满足if条件后,count只会--一次
if(window.get(ch2) < needs.get(ch2)){
count--;
}
}
left++;
}
// res.add(left-1);
}
return res;
}
}LeetCode最优解, 只用了一个哈希表, 本质是一样的
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++;
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;
}
}大佬的极致总结, 图末是他的公众号二维码
