1classSolution {2publicString findReplaceString(String S,int[] indexes, String[] sources, String[] targets) {3StringBuilder res =newStringBuilder();45HashMap<Integer, Integer> map =newHashMap<>();6for(inti = 0;
S.replace(a.first, a.second.first, a.second.second); }returnS; } }; 参考资料: https://leetcode.com/problems/find-and-replace-in-string/ https://leetcode.com/problems/find-and-replace-in-string/discuss/134758/Java-O(n)-solution https://leetcode.com/problems/find-and-replace-in-string...
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing. For example, if we have S = “...
leetcode890---find and replace pattern打败100% 你有一个单词列表 words 和一个模式 pattern,你想知道 words 中的哪些单词与模式匹配。 如果存在字母的排列 p ,使得将模式中的每个字母 x 替换为 p(x) 之后,我们就得到了所需的单词,那么单词与模式是匹配的。 (回想一下,字母的排列是从字母到字母的双射:...
Time Complexity: O(n*m). n = words.length. m is average length of word in words. Space: O(m). regardless res. AC Java: 1classSolution {2publicList<String>findAndReplacePattern(String[] words, String pattern) {3List<String> res =newArrayList<>();4if(pattern ==null|| pattern.length...
var findAndReplacePattern = function(words, pattern) { const norPatter = toNormal(pattern) return words.filter( v => toNormal(v) === norPatter) }; var toNormal = function(string) { let li = string.split('') let tem = []
1. Description Find And Replace in String 2. Solution 解析:Version 1,先排序,根据索引排序,对源字符串和目标字符串也根据索引的排序顺序排序,这样做主要为了判断是否有重叠字符串。遍历所有索引,如果当前索引加上源字符串的长度与下一个索引重叠,则当前索引以及下一个索引对应的字符串都不能替换。如果字符串s根...
classSolution{publicStringfindReplaceString(Strings,int[]indices,String[]sources,String[]targets){StringBuildersb=newStringBuilder();intn=s.length(),m=indices.length,j=0,mark[]=newint[n];for(inti=0;i<m;i++)if(sources[i].equals(s.substring(indices[i],sources[i].length()+indices[i])))...
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