Leetcode0808 有重複字串的排列組合 去重
阿新 • • 發佈:2020-08-07
JAVA 雜湊去重:
final Set<String> reSet = new HashSet<String>(); public final String[] permutation(String S) { search(S, new HashSet<Integer>(), new StringBuilder()); return setToString(reSet); } private final void search(String s, Set<Integer> path, StringBuilder sb) {if (sb.length() == s.length()) { String re = sb.toString(); if (!reSet.contains(re)) { reSet.add(re); } return; } for (int i = 0; i < s.length(); i++) { if (path.contains(i)) { continue; }char currentChar = s.charAt(i); sb.append(currentChar); path.add(i); search(s, path, sb); path.remove(i); sb.deleteCharAt(sb.length() - 1); } } private final String[] setToString(Set<String> set) { String[] reArr= new String[set.size()]; Iterator iterator = set.iterator(); int i = 0; while (iterator.hasNext()) { reArr[i] = (String) iterator.next(); i++; } return reArr; }
JAVA 排序去重:
List<String> reList = new LinkedList<String>(); public final String[] permutation2(String S) { char[] chars = S.toCharArray(); Arrays.sort(chars); search2(chars, new StringBuilder()); return listToArr(reList); } private final void search2(char[] chars, StringBuilder sb) { if (sb.length() == chars.length) { reList.add(sb.toString()); return; } for (int i = 0; i < chars.length; i++) { if (chars[i] == '-' || (i > 0 && chars[i] == chars[i - 1])) { continue; } char currentChar = chars[i]; chars[i] = '-'; search2(chars, sb.append(currentChar)); chars[i] = currentChar; sb.deleteCharAt(sb.length() - 1); } } private final String[] listToArr(List<String> list) { String[] arr = new String[list.size()]; for (int i = 0; i < list.size(); i++) { arr[i] = list.get(i); } return arr; }
JS 排序去重:
/** * @param {string} S * @return {string[]} */ var permutation = function (S) { let sArr = S.split('').sort(); let reArr = []; search(sArr, '', reArr); return reArr; }; var search = function (str, reStr, reArr) { if (str.length == reStr.length) { reArr.push(reStr); return; } for (let i = 0; i < str.length; i++) { if (str[i] == '-' || (i > 0 && str[i] == str[i - 1])) { continue; } let currentChar = str[i]; str[i] = '-'; search(str, reStr + currentChar,reArr); str[i] = currentChar; } }