classSolution {publicString customSortString(String S, String T) {if(S==null||S.length()==0||T==null||T.length()==0){returnnull; } String res= "";for(inti = 0; i<S.length(); i++){//System.out.println(S.charAt(i));while(T.indexOf(S.charAt(i))>=0){ res= res +Cha...
classSolution {public:stringcustomSortString(stringS,stringT) {stringres =""; unordered_map<char,int>m;for(charc : T) ++m[c];for(charc : S) { res+=string(m[c], c); m[c]=0; }for(auto a : m) { res+=string(a.second, a.first); }returnres; } }; 下面这种解法的思路和上...
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 s...
其他排序如 "dbca" 或 "bcda" 也是有效的,只要维持 "b"、"c"、"a" 的顺序。 提示: 1 <= order.length <= 26 1 <= s.length <= 200 order和s由小写英文字母组成 order中的所有字符都不同 排序:最热 1 2 3 4 5 6 classSolution{ public: stringcustomSortString(stringorder,strings) { } };...
解法一(自定义排序): class Solution: def customSortString(self, S: str, T: str) -> str: count = {} for ch in set(S): count[ch] = S.index(ch) return "".join(sorted(T, key=lambda k: count[k] if k in count else -1)) 1 2 3 4 5 6 7版权...
class Solution { public: string customSortString(string S, string T) { map<int,char> m;//S中的字符的出现次序--对应字符 int i; for(i = 0; i < S.size(); ++i) m[i] = S[i]; int count[26] = {0}; for(i = 0; i < T.size(); ++i) count[T[i]-'a']++;//字符个数...
classSolution:defcustomSortString(self,order:str,s:str)->str:ret=""sList=list(s)foriinorder:#print(i)whileTrue:# 利用remove的特性,把對應字母刪增,直到沒法進行為止try:sList.remove(i)ret+=i #print(ret)except:breakret+="".join(sList)returnretif__name__=='__main__':S="cba"T="abc...
char *customSortString(char *s, char *t){ char *result = NULL; int len = strlen(t); int endPos = 0; result = (char *)malloc(len + 1); if (result == NULL) { return NULL; } memset(result, 0, len + 1); memcpy(result, t, len); ...
var customSortString = function(S, T) { let orderArr = S.split('') let countMap = new Map() let resultS = '' for (let i=0, SLen=S.length; i<SLen; i++) { countMap.set(S[i], 0) } for (let i=0, TLen=T.length; i<TLen; i++) { ...
order 的所有字母都是 唯一 的,并且以前按照一些自定义的顺序排序。 对s 的字符进行置换,使其与排序的 order 相匹配。更具体地说,如果在 order 中的字符 x 出现字符 y 之前,那么在排列后的字符串中, x 也应该出现在 y 之前。 返回满足这个性质的 s 的任意一种排列 。 示例1: 输入: order = "cba", ...