标签：only 字母 cas and empty inpu letters 出现 turn
Given a string s and a non-empty string p, find all the start indices of p‘s anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input: s: "cbaebabacd" p: "abc" Output: [0, 6] Explanation: The substring with start index = 0 is "cba", which is an anagram of "abc". The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input: s: "abab" p: "ab" Output: [0, 1, 2] Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". The substring with start index = 1 is "ba", which is an anagram of "ab". The substring with start index = 2 is "ab", which is an anagram of "ab".
题目大意：
给定字符串s和非空字符串p，求在s中p的所有字母异位词子串，返回子串索引。
理 解：
我自己的做法是：每遍历s的一个字符作为起始字符与p比较，结果超时了。时间复杂度O(m*n)。
看别人滑动窗口的做法感觉非常优秀。用unordered_map保存字符串p
初始滑动窗口left=right=0，判断right是否在p中出现，满足则右移right。否则左移left；若滑动窗口的字符串即为p的字母异位词，则记录left；更新left = left + 1,match = match - 1，以及Windows表。
注意：移除当前出现次数未0的字符。
代 码 C++：
class Solution { public: vector<int> findAnagrams(string s, string p) { vector<int> res; if(s=="" || s.length()<p.length()) return res; unordered_map<char,int> need,window; unordered_map<char,int>::iterator itr,itrW; int left=0,right=0,match = 0,j; // p字符串存在need哈希表中 for(int i=0;i<p.size();++i){ itr = need.find(p[i]); if(itr!=need.end()) itr->second++; else need.insert(pair<char,int>(p[i],1)); } while(left<=right && right<s.length()){ itr = need.find(s[right]); if(itr!=need.end()){ // s[right] 出现在need中 itrW = window.find(s[right]); if(itrW!=window.end()){ // s[right] 出现在window中 itrW->second++; if(itrW->second > itr->second){ // s[right]数量过多，从left往后的第一个s[right]重新开始 j = s.find(s[right],left); while(left<=j){ if((window.find(s[left]))->second-- == (need.find(s[left]))->second){ if((window.find(s[left]))->second==0) window.erase(s[left]); match--; } left++; } }else if(itrW->second == itr->second){ // s[right]数量满足条件 match++; } }else{ // s[right] 插入window中 window.insert(pair<char,int>(s[right],1)); if(itr->second == 1) match++; } right++; }else{ // s[right] 未出现在need中，重新开始构建窗口 window.clear(); left = right + 1; right = left; match = 0; } if(match==need.size()){ // 找到符合条件的子串，窗口左指针左移一位 res.push_back(left); match--; if ((window.find(s[left]))->second == 1) window.erase(s[left]); else (window.find(s[left]))->second--; left++; } } return res; } };
运行结果：
执行用时 :68 ms, 在所有 C++ 提交中击败了48.43%的用户
letecode [438] - Find All Anagrams in a String
标签：only 字母 cas and empty inpu letters 出现 turn
原文地址：https://www.cnblogs.com/lpomeloz/p/11099517.html
