leetcode: Find All Anagrams in a String
2016-10-28 10:59
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class Solution { public: vector<int> KMP_next(string match) { vector<int>next(match.size()); next[0] = -1; int j = 0, k = -1; while (j<match.size() - 1) { while (k >= 0 && match[k] != match[j]) { k = next[k]; } k++, j++; if (match[j] != match[k])next[j] = k; else { next[j] = next[k];} } return next; } vector<int> findIndexOfOne(string s, string x) { int len = x.size(); vector<int>next = KMP_next(x); vector<int>res; int i = 0; int j = 0; while (i < s.size() ) { while (j >= 0 && x[j] != s[i]) { j = next[j]; } i++, j++; if (j == len) //这个地方对前者的改进,能够输出所有的下标 { res.push_back(i - len); if (i == s.size())break; j = 0; i =i- len+1; } } return res; } void getFullPerMutation(vector<string>&res, string&p,int begin) { int len = p.size(); if (begin >= len) { res.push_back(p); return; } for (int i=begin;i<len;++i) { if (i>begin&&i > 0 && p[i] == p[i - 1])continue; swap(p[begin], p[i]); getFullPerMutation(res, p, begin + 1); swap(p[begin], p[i]); } } vector<int> findAnagrams(string s, string p) { //全排列+改良KMP sort(p.begin(), p.end()); vector<string>permutationOfp; getFullPerMutation(permutationOfp, p,0); vector<int>res; for (string x : permutationOfp) { auto ind = findIndexOfOne(s, x); if (ind.size() > 0) //此处应该可以用insert函数改良 for (int n:ind) { res.push_back(n); } } return res; } };
Memory Limited Exceeded
虽然这种方法失败了,但是中间的基本功都通过了
对string进行全排列,并且考虑去重
next数组的优化
KMP查找可能出现的所有下标,返回vector< int >,如果size为零表示没有查找到
class Solution { public: vector<int> findAnagrams(string s, string p) { vector<int>res; if (s.size() == 0 || p.size() == 0)return res; vector<int>hashInd(128, 0); for (char c : p)hashInd[c]++; int left = 0, right = 0, count = p.size(); while (right < s.size()) { if (hashInd[s[right++]]-- >= 1)count--; if (count == 0)res.push_back(left); if (right - left == p.size() && hashInd[s[left++]]++ >= 0)count++; } return res; } };
这种巧妙地算法可以看做“滑动窗口slide window”
初始化:用一个hash存储char-value,value表示在p中这个char出现的次数,表示这个char还有多少个没有发生配对。另外用一个count=p.size()表示总共需要检测char数量
检测期分为两个状态
第一个:right-left < p.size() 这个时候窗口还没有达到最大size,right每次往后移一个单位,如果s[right]对应的value>=1表示在p中出现了且还没有发生配对,count–,如果s[right]对应的value<1表示当前这个char要么不是p中的,要么已经配对过了
第二个:right-left==p.size() 这个时候窗口达到最大size,right和left同时往后移动一个单位,left往前移动一位,相当于最末char离开窗口,那么hash[char]++, 如果这个char和p发生过有效匹配,count++;
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