532. K-diff Pairs in an Array
Given an array of integers and an integer
k, you need to find the number of
unique k-diff pairs in the array. Here a
k-diff pair is defined as an integer pair (i, j), where
i and
j are both numbers in the array and their
absolute difference is
k.
Example 1:
Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:
Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:
Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Note:
- The pairs (i, j) and (j, i) count as the same pair.
- The length of the array won't exceed 10,000.
- All the integers in the given input belong to the range: [-1e7, 1e7].
public class Solution {
public int findPairs(int[] nums, int k) {
if (k < 0) return 0;
Set<Integer> set = new HashSet<>();
Set<Integer> firtInPair = new HashSet<>();
for (int i = 0; i < nums.length; i++) {
int cur = nums[i];
if (set.contains(cur + k)) {
firstInPair.add(cur);
}
if (set.contains(cur - k)) {
firstInPair.add();
}
set.add(cur);
}
return cur.size();
}
}
438. Find All Anagrams in a String
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".
Sliding window
另一种写法是用array［256］来代替Map，代码会更简洁。因为array[char] == 0可以同时代表没有出现过的char和已经用完的char
public class Solution {
public List findAnagrams(String s, String p) {
Map occurance = getOccurance(p);
return findIndexes(s, p.length(), occurance, new HashMap<>(occurance));
}
private Map getOccurance(String p) {
Map occ = new HashMap<>();
for (int i = 0; i < p.length(); i++) {
char c = p.charAt(i);
if (occ.containsKey(c)) {
occ.put(c, occ.get(c) + 1);
} else {
occ.put(c, 1);
}
}
return occ;
}
private List findIndexes(String s, int count, Map occ, Map backup) {
List rt = new ArrayList<>();
int start = 0;
for (int i = 0; i < s.length(); i++) {
char c =s.charAt(i);
if (!occ.containsKey(c)) {
// reset
occ = new HashMap<>(backup);
start = i + 1;
count = occ.size();
continue;
}
while (occ.get(c) == 0) {
occ.put(s.charAt(start), occ.get(s.charAt(start)) + 1);
start++;
count++;
}
count--;
if (count == 0) {
rt.add(start);
}
occ.put(c, occ.get(c) - 1);
}
return rt;
}
}
387. First Unique Character in a String
Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
Examples:
s = "leetcode"
return 0.
s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.
public class Solution {
public int firstUniqChar(String s) {
int occ[] = new int[256];
for (int i = 0; i < s.length(); i++) {
int val = s.charAt(i) - '0';
occ[val]++;
}
for (int i = 0; i < s.length(); i++) {
int val = s.charAt(i) - '0';
if (occ[val] == 1) {
return i;
}
}
return -1;
}
}
459. Repeated Substring Pattern
Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.
Example 1:
Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.
Example 2:
Input: "aba"
Output: False
Example 3:
Input: "abcabcabcabc"
Output: True
Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)
可以用KMP
以下方法可以稍微再简化：inner loop里可以把所有的substring加起来然后跟原来的比较
public class Solution {
public boolean repeatedSubstringPattern(String s) {
for (int i = 0; i < s.length() / 2; i++) {
if (s.length() % (i + 1) != 0) {
continue;
}
String sub = s.substring(0, i + 1);
boolean flag = true;
for (int j = i + 1; j < s.length() - i; j += i + 1) {
String secSub = s.substring(j, j + i + 1);
if (!secSub.equals(sub)) {
flag = false;
break;
}
}
if (flag == true) return true;
}
return false;
}
}
