求字符串内不包含重复字符的最长子串的集合

来源：互联网发布：在计算机中算法是指编辑：程序博客网时间：2024/06/05 14:20

方法一：循环遍历，采用Set保存数据，复杂度O(n^2)

private static Set<String> getSubList(String str) {if (null == str || str.trim().length() == 0) {return null;}Set<String> result = new HashSet<String>();int maxLength = 0;StringBuilder sb = new StringBuilder();char[] array = str.toCharArray();int length = array.length;for (int i = 0; i < length; i++) {int j = i + 1;sb.delete(0, sb.length());sb.append(array[i]);while ((length - j) >= maxLength) {sb.delete(0, sb.length());for (; j < length; j++) {if (sb.toString().contains(String.valueOf(array[j]))) {break;} else {sb.append(String.valueOf(array[j]));}}//System.out.println(j + "," + sb.toString());if (sb.length() > maxLength) {maxLength = sb.length();result.clear();result.add(sb.toString());} else if (sb.length() == maxLength) {result.add(sb.toString());}}}return result;}

方法2：

采用一个map来做中间缓存，用list保存结果，map的key存储的是字符，value存储的是该字符当前的位置，首先设置一个当前current位置，默认从0开始，那么从开始遍历字符串，如果map当中不包含这个字符，那么用这个字符当前所在的位置减去current位置，然后与最大长度做比较，选大的成为最大长度，然后把当前字符的以及位置放入map，同时把从current位置的字符到current+maxLength位置的字符组成子串放入一个list中，保存前需要判断当前list中已经保存的字符的长度是否大于将要保存的字符长度，小于则将list清除，复杂度 O(n)

public static List<String> getSubList2(String str){if (null == str || str.trim().length() == 0) {return null;}Map<Character,Integer> tempMap = new HashMap<>();List<String> result = new ArrayList<String>();StringBuilder sb = new StringBuilder();int maxLength = 0;char[] array = str.toCharArray();int length = array.length;int current = 0;for(int i=0;i<length;i++){if(tempMap.containsKey(array[i])){current = Math.max(tempMap.get(array[i])+1,current);//System.out.println((i-current+1)+","+maxLength);if((i-current+1)>maxLength){maxLength = i-current+1;sb.delete(0, sb.length());for(int j=current;j<current+maxLength;j++){sb.append(array[j]);}if(result.size() > 0){if(result.get(0).length()<maxLength){result.clear();}}result.add(sb.toString());}else if((i-current+1) == maxLength){sb.delete(0, sb.length());for(int j=current;j<current+maxLength;j++){sb.append(array[j]);}if(result.size() > 0){if(result.get(0).length()<maxLength){result.clear();}}result.add(sb.toString());}}else{//System.out.println((i-current+1)+","+maxLength);if((i-current+1)>maxLength){maxLength = i-current+1;sb.delete(0, sb.length());for(int j=current;j<current+maxLength;j++){sb.append(array[j]);}if(result.size() > 0){if(result.get(0).length()<maxLength){result.clear();}}result.add(sb.toString());}else if((i-current+1) == maxLength){sb.delete(0, sb.length());for(int j=current;j<current+maxLength;j++){sb.append(array[j]);}if(result.size() > 0){if(result.get(0).length()<maxLength){result.clear();}}result.add(sb.toString());}}tempMap.put(array[i], i);}//System.out.println(maxLength);return result;}

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