1002. Find Common Characters
Given a string array words, return an array of all characters that show up in all strings within the words (including duplicates) . You may return the answer in any order.
Example 1:
Input: words = ["bella","label","roller"]
Output: ["e","l","l"]
Example 2:
Input: words = ["cool","lock","cook"]
Output: ["c","o"]
这道题虽然是一道简单题，但是比较难想到解法。先统计第一个字符串所有字符出现的次数，代码如下：
int[] hash = new int[26]; // 用来统计所有字符串里字符出现的最小频率
for (int i = 0; i < A[0].size(); i++) { // 用第一个字符串给hash初始化
hash[A[0][i] - 'a']++;
}
接下来，把其他字符串里字符的出现次数也统计出来一次放在hashOtherStr中。
然后hash 和 hashOtherStr 取最小值，这是本题关键所在，此时取最小值，就是 一个字符在所有字符串里出现的最小次数了。
代码如下：
int hashOtherStr[26] = {0}; // 统计除第一个字符串外字符的出现频率
for (int i = 1; i < A.size(); i++) {
int[] hashOtherStr = new int[26];
for (int j = 0; j < A[i].size(); j++) {
hashOtherStr[A[i][j] - 'a']++;
}
// 这是关键所在
for (int k = 0; k < 26; k++) { // 更新hash，保证hash里统计26个字符在所有字符串里出现的最小次数
hash[k] = Math.min(hash[k], hashOtherStr[k]);
}
}
最后的代码：
class Solution {
public List<String> commonChars(String[] words) {
List<String> res = new ArrayList<String>();
int[] hash = new int[26];
if (words.length == 0) {
return res;
}
for (int i = 0; i < words.length; i++){
String str = words[i];
if (i == 0) {
for (int j = 0; j < str.length(); j++){
hash[str.charAt(j) - 'a']++;
}
} else {
int[] otherhash = new int[26];
for (int j = 0; j < str.length(); j++){
otherhash[str.charAt(j) - 'a'] ++;
}
for(int k=0; k < 26; k++){
hash[k] = Math.min(hash[k], otherhash[k]);
}
}
}
for(int i=0; i < 26; i++){
while(hash[i] != 0){
char c= (char) (i+'a');
res.add(String.valueOf(c));
hash[i]--;
}
}
return res;
}
}
Write an algorithm to determine if a number n is happy.
A happy number is a number defined by the following process:
- Starting with any positive integer, replace the number by the sum of the squares of its digits.
- Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
- Those numbers for which this process ends in 1 are happy.
Return true if n is a happy number, and false if not.
Example 1:
Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Example 2:
Input: n = 2
Output: false
首先题里提醒了如果不等于1会陷入无限循环，所以可以通过哈希表存储
从一个integer n获取这个integer每一位的digit可以通过n%10，然后再把n/10直到n为0, 见下面的getNextNumber
class Solution {
public boolean isHappy(int n) {
Map<Integer, Integer> map = new HashMap<>();
while(n != 1 && !map.containsKey(n)) {
map.put(n, 1);
n = getNextNumber(n);
}
return n == 1;
}
public int getNextNumber(int n) {
int res = 0;
while(n > 0) {
int temp = n % 10;
res += temp * temp;
n = n/10;
}
return res;
}
}
