剑指Offer 38 字符串的排列
作者:互联网
JAVA 回溯解法:
public final String[] permutation(String s) { Set<String> reSet = new HashSet<String>(); search(s, 0, reSet, new StringBuilder(), new HashSet<Integer>()); return reSet.toArray(new String[reSet.size()]); } private final void search(String s, int len, Set<String> reSet, StringBuilder sb, Set<Integer> path) { if (len == s.length()) { String reStr = sb.toString(); if (!reSet.contains(reStr)) { reSet.add(reStr); } return; } for (int i = 0; i < s.length(); i++) { if (path.contains(i)) { continue; } char currentChar = s.charAt(i); path.add(i); sb.append(currentChar); int nextLen = len + 1; search(s, nextLen, reSet, sb, path); path.remove(i); sb.deleteCharAt(len); } }
JS 回溯解法:
var permutation = function (s) { let reSet = new Set(); search(s, 0, reSet, new Array(s.length), ""); return Array.from(reSet); }; var search = function (s, len, reSet, path, reStr) { if (len == s.length) { if (!reSet.has(reStr)) { reSet.add(reStr); } return; } let nextLen = len + 1; for (let i = 0; i < s.length; i++) { if (path[i] === 1) { continue; } let currentChar = s.charAt(i); path[i] = 1; search(s, nextLen, reSet, path, reStr + currentChar); path[i] = undefined; } }
标签:reSet,search,38,Offer,reStr,len,new,字符串,path 来源: https://www.cnblogs.com/niuyourou/p/13420028.html