标签:return find tin 区分 null string 重复 基本 i++
我就想了个递归, 还是没有区分掉一些重复的情况,worst case O(2^n)基本同暴力解
Map<Integer, Set<String>> allSubSet = new HashMap();
Set<String> getAllPalidrome(String s, int x, int y){
int ind = x * s.length() + y;
if(allSubSet.constainsKey(ind)) return allSubSet.get(ind);
Set<String> ret = new HashSet();
if (s == null || s.size() == 0) { ret.add(""); return ret;}
if (s.size() == 1) { ret.add(s); return ret;}
for(int i = x; i <= y; i++){
for (int j = y; j >= i; j--){
if (s.charAt(i) == s.charAt(j)){
Set<String> subSet = getAllPalidrome(s, i + 1, j - 1);
ret.addAll(subSet);
for (String str : subSet) ret.add(s.charAt(i) + str + s.charAt(i));
}
}
}
allSubSet.put(ind, ret);
return ret;
}
find all palidrome string by deleting any letter from the given string
标签:return find tin 区分 null string 重复 基本 i++
原文地址:http://www.cnblogs.com/apanda009/p/7965312.html