## Monday, April 23, 2018

### LeetCode 791 - Custom Sort String

https://leetcode.com/problems/custom-sort-string/description/
`S` and `T` are strings composed of lowercase letters. In `S`, no letter occurs more than once.
`S` was sorted in some custom order previously. We want to permute the characters of `T` so that they match the order that `S` was sorted. More specifically, if `x` occurs before `y` in `S`, then `x` should occur before `y` in the returned string.
Return any permutation of `T` (as a string) that satisfies this property.
```Example :
Input:
S = "cba"
T = "abcd"
Explanation:
"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a".
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.
```

Note:
• `S` has length at most `26`, and no character is repeated in `S`.
• `T` has length at most `200`.
• `S` and `T` consist of lowercase letters only.
https://blog.csdn.net/Shauna_Wu/article/details/79381747
最先考虑的就是去储存S中字符的顺序，然后对T进行排序。但是顺序存储会十分复杂，所以反过来进行。先对要排序的字符串T进行遍历，得出每个字符各自出现的次数。然后依照S中的顺序，把字符按序排列，最后加上没有排序要求的字符即可。
public String customSortString(String S, String T) { int[] count = new int[26]; StringBuffer s = new StringBuffer(); for(char c : T.toCharArray()){ count[c - 'a']++; } for(char c : S.toCharArray()){ while(count[c - 'a'] > 0){ s.append(c); count[c - 'a']--; } } for(char c = 'a'; c <= 'z'; c++){ while(count[c - 'a'] > 0){ s.append(c); count[c - 'a']--; } } return s.toString(); }

public String customSortString(String S, String T) {
if (S == null || T == null) {
return T;
}

Map<Character, Long> freqs = getFreq(T);
StringBuilder sb = new StringBuilder();
for (Character ch : S.toCharArray()) {
if (freqs.containsKey(ch)) {
append(sb, ch, freqs.get(ch));
freqs.remove(ch);
} // else do nothing
}

for (Entry<Character, Long> entry : freqs.entrySet()) {
append(sb, entry.getKey(), entry.getValue());
}

return sb.toString();
}

private void append(StringBuilder sb, Character ch, long count) {
for (int i = 0; i < count; i++) {
sb.append(ch);
}
}

private Map<Character, Long> getFreq(String T) {
return T.chars().mapToObj(ch -> (char) ch)
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
// Stream<char[]> s = Stream.of(T.toCharArray());
// .collect(Collectors.groupingBy(o->o));

}
Solution 1: HashTable + Sorting
1. Store the order of char in a hashtable
2. Sort the string based on the order
Time complexity: O(nlogn)
Space complexity: O(128)
string customSortString(string S, string T) {
vector<int> order(128, INT_MAX);
int i = 0;
for (char c: S)
if (order[c] == INT_MAX) order[c] = ++i;
std::sort(T.begin(), T.end(), [&order](const char c1, const char c2){
return order[c1] < order[c2];
});
return T;
}

https://www.cnblogs.com/pk28/p/8470187.html
string customSortString(string S, string T) { int n = T.size(); vector<int> vis(n); string ans = ""; for (int i = 0; i < S.size(); ++i) { for (int j = 0; j < n; ++j) { if (!vis[j] && S[i] == T[j]) { ans += T[j]; vis[j] = 1; } } } for (int i = 0; i < T.size(); ++i) { if (!vis[i]) ans += T[i]; } return ans; }