标签：bsp result new Matter har null rac 操作 png
滑动窗口（sliding windows algorithm）这种方法，专门用于解决区间解的问题。它在运算的时候，将解集放在窗口中，结束的时候比对是否符合预期。在运算的过程中，会对窗口的左右边缘进行操作（扩大、缩小）。特别针对于线性输入解决，滑动窗口就很形象了。
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
题意是从一串输入字符串s中，找到能全包含words数组的起点（数组元素等长），这个起点可能有多个，而且不需要关心顺序。
Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively. The output order does not matter, returning [9,0] is fine too.
比如这个例子，他在第0个位置和第9个位置可以包含words数组，也就是0位置和9位置各有一个长度为6的窗口可以囊括words数组。
每次扩展使用单词列表中的词长，如果扩展过程中不符合预期则清除窗口，窗口左端在当前词总数大于词组总数的时候，计算总数是单词的长度（因为s除以单词长度余数是0到单词长度之间，覆盖了余数就能覆盖整个场景）。
public class FindSubstring { public void test(){ // [0,9] String s1 = "barfoothefoobarman"; String[] words1 = {"foo","bar"}; System.out.println(findSubstring(s1,words1)); // [] String s2 = "wordgoodgoodgoodbestword"; String[] words2 = {"word","good","best","word"}; System.out.println(findSubstring(s2,words2)); } /** * 单词等长 */ public List<Integer> findSubstring(String s, String[] words) { List<Integer> result = new ArrayList<>(); if(s == null || words.length == 0){ return result; } int step = words[0].length(); Map<String,Integer> counter = new HashMap<>(); for(String word :words){ counter.merge(word, 1, (a, b) -> a + b); } for(int i=0;i<step;++i){ Map<String,Integer> window = new HashMap<>(); int left = i; int right = i; while (right <= s.length() - step && left <= s.length() - step*words.length){ String sub = s.substring(right,right + step); window.merge(sub,1,(a , b)->a + b); if(!counter.containsKey(sub)){ window.clear(); right += step; left = right; continue; } while (window.get(sub) > counter.get(sub)){ String drop = s.substring(left,left+step); Integer dec = window.get(drop); if(dec != null){ if(dec<=1){ window.remove(drop); }else { window.put(drop,dec-1); } } left += step; } right += step; if(right - left == step * words.length){ result.add(left); } } } return result; } }
标签：bsp result new Matter har null rac 操作 png
原文地址：https://www.cnblogs.com/chentingk/p/11865392.html
