https://leetcode.com/problems/reconstruct-original-digits-from-english/
https://discuss.leetcode.com/topic/63386/one-pass-o-n-java-solution-simple-and-clear
for zero, it's the only word has letter 'z',
for two, it's the only word has letter 'w',
......
so we only need to count the unique letter of each word, Coz the input is always valid.
http://bookshadow.com/weblog/2016/10/16/leetcode-reconstruct-original-digits-from-english/
https://discuss.leetcode.com/topic/63486/short-matrix-solution
https://discuss.leetcode.com/topic/64207/java-naive-solution
Given a non-empty string containing an out-of-order English representation of digits
0-9
, output the digits in ascending order.
Note:
- Input contains only lowercase English letters.
- Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
- Input length is less than 50,000.
Example 1:
Input: "owoztneoer" Output: "012"
Example 2:
Input: "fviefuro" Output: "45"X.
https://discuss.leetcode.com/topic/63386/one-pass-o-n-java-solution-simple-and-clear
for zero, it's the only word has letter 'z',
for two, it's the only word has letter 'w',
......
so we only need to count the unique letter of each word, Coz the input is always valid.
public String originalDigits(String s) {
int[] count = new int[10];
for (int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if (c == 'z') count[0]++;
if (c == 'w') count[2]++;
if (c == 'x') count[6]++;
if (c == 's') count[7]++; //7-6
if (c == 'g') count[8]++;
if (c == 'u') count[4]++;
if (c == 'f') count[5]++; //5-4
if (c == 'h') count[3]++; //3-8
if (c == 'i') count[9]++; //9-8-5-6
if (c == 'o') count[1]++; //1-0-2-4
}
count[7] -= count[6];
count[5] -= count[4];
count[3] -= count[8];
count[9] = count[9] - count[8] - count[5] - count[6];
count[1] = count[1] - count[0] - count[2] - count[4];
StringBuilder sb = new StringBuilder();
for (int i = 0; i <= 9; i++){
for (int j = 0; j < count[i]; j++){
sb.append(i);
}
}
return sb.toString();
}
https://discuss.leetcode.com/topic/63382/share-my-simple-and-easy-o-n-solution public String originalDigits(String s) {
if(s==null || s.length()==0) return "";
int[] count = new int[128];
for(int i=0;i<s.length();i++) count[s.charAt(i)]++;
int[] num = new int[10];
num[0] = count['z'];
num[2] = count['w'];
num[6] = count['x'];
num[8] = count['g'];
num[7] = count['s']-count['x'];
num[5] = count['v']-count['s']+count['x'];
num[4] = count['u'];
num[3] = count['h']-count['g'];
num[1] = count['o']-count['z']-count['w']-count['u'];
num[9] = count['i']-count['x']-count['g']-count['v']+count['s']-count['x'];
String ret = new String();
for(int i=0;i<10;i++)
for(int j=num[i];j>0;j--) ret += String.valueOf(i);
return ret;
}
http://bookshadow.com/weblog/2016/10/16/leetcode-reconstruct-original-digits-from-english/
字符统计 + 枚举
def originalDigits(self, s):
"""
:type s: str
:rtype: str
"""
cnts = collections.Counter(s)
nums = ['six', 'zero', 'two', 'eight', 'seven', 'four', 'five', 'nine', 'one', 'three']
numc = [collections.Counter(num) for num in nums]
digits = [6, 0, 2, 8, 7, 4, 5, 9, 1, 3]
ans = [0] * 10
for idx, num in enumerate(nums):
cntn = numc[idx]
t = min(cnts[c] / cntn[c] for c in cntn)
ans[digits[idx]] = t
for c in cntn:
cnts[c] -= t * cntn[c]
return ''.join(str(i) * n for i, n in enumerate(ans))https://discuss.leetcode.com/topic/63486/short-matrix-solution
https://discuss.leetcode.com/topic/64207/java-naive-solution
public String originalDigits(String s) {
char[] dic = {'z','w','g','x','u','s','v','o','t','i'};
String[] digits = {"zero","two","eight","six","four","seven","five","one","three","nine"};
int[] index = {0,2,8,6,4,7,5,1,3,9};
int[] map = new int[26];
int[] arr = new int[10];
for(int i=0; i<s.length(); i++){
map[s.charAt(i)-'a']++;
}
for(int i=0; i<10; i++){
getNum(map, dic[i], digits[i], arr, index[i]);
}
StringBuilder sb = new StringBuilder();
for(int i=0; i<10; i++){
for(int t=0; t<arr[i]; t++){
sb.append(i);
}
}
return sb.toString();
}
private void getNum(int[] map, char c, String s, int[] arr, int index){
int dup = map[c-'a'];
for(int i=0; i<dup; i++){
arr[index]++;
for(int j=0; j<s.length(); j++){
map[s.charAt(j)-'a']--;
}
}
}